If ${\rm{\theta }} = {\sin ^{ - 1}}x + {\cos ^{ - 1}}x - {\tan ^{ - 1}}x \ge 0,$ then the smallest interval in which ${\rm{\theta }}$ lies, is given by

$\frac{\pi }{4} \le {\rm{\theta }} \le \frac{\pi }{2}$

$\frac{\pi }{2} \le {\rm{\theta }} \le \frac{{3\pi }}{4}$

$ - \frac{\pi }{4} \le {\rm{\theta }} \le 0$

$0 \le {\rm{\theta }} \le \frac{\pi }{4}$

If ${\rm{\alpha }} = {\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{1}{3}{\rm{and}}\;{\rm{\beta }} = {\cos ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{1}{3},\;$ then

${\rm{\alpha }} = {\rm{\beta }}$

${\rm{\alpha }} > {\rm{\beta }}$

If ${\rm{\alpha }} = {\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{1}{3}{\rm{and}}\;{\rm{\beta }} = {\cos ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{1}{3},\;$ then

${\rm{\alpha }} = {\rm{\beta }}$

${\rm{\alpha }} > {\rm{\beta }}$

INVERSE TRIGONOMETRY MCQs for JEE MATHEMATICS — Mixed difficulty, Page 6

151 Easy

151 of 581

Application +4 | -1

\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)

is equal to

View answer and explanation

Correct option: B

-\frac{\pi}{3}

Option B (

-\frac{{\pi}}{{3}}

) is correct because it is the result of subtracting the principal values of the inverse secant from the inverse tangent.

Solution approach

Correct Answer + Core Reason: $-\frac{{\pi}}{{3}}$ . The expression is evaluated by finding the principal values of the inverse tangent and inverse secant functions and then subtracting them.

Step-by-Step Solution:

Step 1: $\tan^{-1}(\sqrt{3}) = \frac{{\pi}}{{3}}$ .

Step 2: $\sec^{-1}(-2) = \pi - \frac{{\pi}}{{3}} = \frac{{2\pi}}{{3}}$ .

Step 3: Therefore, $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2) = \frac{{\pi}}{{3}} - \frac{{2\pi}}{{3}} = -\frac{{\pi}}{{3}}$ .

Why other options are wrong:
- Option A: $\pi$ is incorrect because the subtraction is not performed correctly.
- Option C: $\frac{{\pi}}{{3}}$ is incorrect because the subtraction is not performed correctly.
- Option D: $\frac{{2\pi}}{{3}}$ is incorrect because the subtraction is not performed correctly.

Key Concepts:
- Inverse trigonometric functions
- Principal values of inverse trigonometric functions

Memory Aids:
- Remember the range of the principal value of $\sec^{-1}x$ is $[0, \pi] - \{\frac{{\pi}}{{2}}\}$ .

Quick Tips:
- Use the definition of the inverse trigonometric function and its range to find the principal value.

152 Medium

152 of 581

Application +4 | -1

The principal value of

\cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left[\cos \left(\frac{9 \pi}{10}\right)-\sin \left(\frac{9 \pi}{10}\right)\right]\right\}

is

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the principal value of the given expression simplifies to

\frac{{17\pi}}{{20}}

, which is not among the provided options.

Solution approach

Correct Answer + Core Reason: None of these. The correct principal value is $\frac{{17\pi}}{{20}}$ , which is not in the options.

Step-by-Step Solution:

Step 1: Simplify the expression inside the $\cos^{-1}$ function.
\frac{{1}}{{ $\sqrt{{2}}$ }} \left[ \cos \left( $\frac{{9π}}{{10}}$ \right) - \sin \left( $\frac{{9π}}{{10}}$ \right) \right] = \frac{{1}}{{ $\sqrt{{2}}$ }} \cos \left( $\frac{{9π}}{{10}}$ \right) - \frac{{1}}{{ $\sqrt{{2}}$ }} \sin \left( $\frac{{9π}}{{10}}$ \right) $.\n\n**Step 2:** Rewrite using$ \cos( $\frac{π}{4}$ ) = \sin( $\frac{π}{4}$ ) = \frac{{1}}{{ $\sqrt{{2}}$ }} = \cos( $\frac{π}{4}$ )\cos( $\frac{{9π}}{{10}}$ ) - \sin( $\frac{π}{4}$ )\sin( $\frac{{9π}}{{10}}$ ) $.\n\n**Step 3:** Use the cosine addition formula:$ \cos(A + B) = \cos A \cos B - \sin A \sin B $. So, the expression becomes$ \cos( $\frac{π}{4}$ + $\frac{{9π}}{{10}}$ ) = \cos( $\frac{{5π + 18π}}{{20}}$ ) = \cos( $\frac{{23π}}{{20}}$ ) $.\n\n**Step 4:** Now we have$ \cos^-1 \left[ \cos \left( $\frac{{23π}}{{20}}$ \right) \right] $. Since$ $\frac{{23π}}{{20}}$ $is not in the range$ [0, π] $, we need to find an equivalent angle.\n\n**Step 5:**$ $\frac{{23π}}{{20}}$ = 2π - $\frac{{17π}}{{20}}$ $. So,$ \cos( $\frac{{23π}}{{20}}$ ) = \cos(2π - $\frac{{17π}}{{20}}$ ) = \cos(- $\frac{{17π}}{{20}}$ ) = \cos( $\frac{{17π}}{{20}}$ ) $.\n\n**Step 6:** Therefore,$ \cos^-1 \left[ \cos \left( $\frac{{23π}}{{20}}$ \right) \right] = \cos^-1 \left[ \cos \left( $\frac{{17π}}{{20}}$ \right) \right] = $\frac{{17π}}{{20}}$ $.\n\n**Why other options are wrong:**\n- **Option A ($ $\frac{{3π}}{{20}}$ $):** This is incorrect because the correct principal value is$ $\frac{{17π}}{{20}}$ $.\n- **Option B ($ $\frac{{7π}}{{20}}$ $):** This is incorrect because the correct principal value is$ $\frac{{17π}}{{20}}$ $.\n- **Option C ($ $\frac{{7π}}{{10}}$ $):** This is incorrect because the correct principal value is$ $\frac{{17π}}{{20}}$ $.\n\n**Key Concepts:**\n- Inverse trigonometric functions\n- Cosine addition formula\n- Principal value of inverse cosine function\n\n**Memory Aids:**\n- Remember the cosine addition formula:$ \cos(A + B) = \cos A \cos B - \sin A \sin B $.\n- Remember the range of$ \cos^-1 x $is$ [0, π] $.\n\n**Quick Tips:**\n- Simplify the expression inside the inverse trigonometric function first.\n- Find an equivalent angle within the principal range.$

153 Easy

153 of 581

Application +4 | -1

The value of

x

for which

{\cos ^{ - 1}}(\cos 4) > 3{x^2} - 4x\;

is

View answer and explanation

Correct option: D

\left( {\frac{{2 - \sqrt {6\pi - 8} }}{3},\frac{{2 + \sqrt {6\pi - 8} }}{3}} \right)

Option D (

\left( {\frac{{2 - \sqrt {6\pi - 8} }}{3},\frac{{2 + \sqrt {6\pi - 8} }}{3}} \right)

) is correct because it represents the interval of x values that satisfy the inequality after correctly evaluating

{\cos ^{ - 1}}(\cos 4)

.

Solution approach

Correct Answer + Core Reason: $\left( {\frac{{2 - \sqrt {6\pi - 8} }}{3},\frac{{2 + \sqrt {6\pi - 8} }}{3}} \right)$ . The interval is obtained by solving the inequality after simplifying the inverse cosine term.

Step-by-Step Solution:

Step 1: Evaluate ${\cos ^{ - 1}}(\cos 4)$ . Since $4$ is not in the range $[0, \pi]$ , we need to find an equivalent angle in that range. $2\pi - 4$ is in the range $[0, \pi]$ , and $\cos(2\pi - 4) = \cos(4)$ . Therefore, ${\cos ^{ - 1}}(\cos 4) = 2\pi - 4$ .

Step 2: Substitute into the inequality: $2\pi - 4 > 3{x^2} - 4x$ .

Step 3: Rearrange the inequality: $3{x^2} - 4x - (2\pi - 4) < 0$ .

Step 4: Solve the quadratic inequality. The roots of the quadratic equation $3x^2 - 4x - (2\pi - 4) = 0$ are given by $x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \frac{{4 \pm \sqrt {16 + 12(2\pi - 4)} }}{6} = \frac{{4 \pm \sqrt {16 + 24\pi - 48} }}{6} = \frac{{4 \pm \sqrt {24\pi - 32} }}{6} = \frac{{2 \pm \sqrt {6\pi - 8} }}{3}$ .

Step 5: Since the quadratic has a positive leading coefficient, the inequality is satisfied between the roots: $\frac{{2 - \sqrt {6\pi - 8} }}{3} < x < \frac{{2 + \sqrt {6\pi - 8} }}{3}$ .

Why other options are wrong:
- Option A: $\left( {0,\frac{{2 + \sqrt {6\pi - 8} }}{3}} \right)$ : This is incorrect because it only provides one bound of the interval and misses the lower bound.
- Option B: $\left( {\frac{{2 - \sqrt {6\pi - 8} }}{3},0} \right)$ : This is incorrect because it only provides one bound of the interval and misses the upper bound.
- Option C: $\left( { - 2,2} \right)$ : This is incorrect because it does not accurately represent the solution to the inequality.

Key Concepts:
- Inverse trigonometric functions
- Quadratic inequalities

Memory Aids:
- Remember that ${\cos ^{ - 1}}(\cos x) = x$ only if $x$ is in the range $[0, \pi]$

154 Easy

154 of 581

Application +4 | -1

\cot ^{-1}\left(-\frac{1}{2}\right)+\cot ^{-1}\left(-\frac{1}{3}\right)

is equal to

View answer and explanation

Correct option: B

\frac{5 \pi}{4}

Option B (

\frac{5 \pi}{4}

) is correct because it is the value obtained by applying the inverse cotangent addition formula and considering the range of the inverse cotangent function.

Solution approach

Correct Answer + Core Reason: $\frac{5 \pi}{4}$ . Using the inverse cotangent addition formula and considering the range.

Step-by-Step Solution:

Step 1: We need to find the value of $\cot^{-1}(-\frac{1}{2}) + \cot^{-1}(-\frac{1}{3})$ .

Step 2: Let $x = \cot^{-1}(-\frac{1}{2})$ and $y = \cot^{-1}(-\frac{1}{3})$ . Then $\cot x = -\frac{1}{2}$ and $\cot y = -\frac{1}{3}$ .

Step 3: We know that $\cot^{-1}(-a) = \pi - \cot^{-1}(a)$ . So, $x = \pi - \cot^{-1}(\frac{1}{2})$ and $y = \pi - \cot^{-1}(\frac{1}{3})$ .

Step 4: Therefore, $\cot^{-1}(-\frac{1}{2}) + \cot^{-1}(-\frac{1}{3}) = \pi - \cot^{-1}(\frac{1}{2}) + \pi - \cot^{-1}(\frac{1}{3}) = 2\pi - [\cot^{-1}(\frac{1}{2}) + \cot^{-1}(\frac{1}{3})]$ .

Step 5: We also know that $\cot^{-1}(a) = \tan^{-1}(\frac{1}{a})$ . So, $\cot^{-1}(\frac{1}{2}) = \tan^{-1}(2)$ and $\cot^{-1}(\frac{1}{3}) = \tan^{-1}(3)$ .

Step 6: Therefore, $\cot^{-1}(-\frac{1}{2}) + \cot^{-1}(-\frac{1}{3}) = 2\pi - [\tan^{-1}(2) + \tan^{-1}(3)]$

155 Easy

155 of 581

Application +4 | -1

A root of the equation

17 x^2+17 x \tan \left[2 \tan ^{-1}(1 / 5)-\pi / 4\right]-10=0

is

View answer and explanation

Correct option: D

1

Option D (1) is correct because simplifying the trigonometric expression and solving the quadratic equation yields 1 as a root.

Solution approach

Correct Answer + Core Reason: 1. Simplifying the trigonometric expression and solving the quadratic equation.

Step-by-Step Solution:

Step 1: Given the equation $17x^2 + 17x \tan \left[2 \tan^{-1} (1/5) - \pi/4\right] - 10 = 0$ .

Step 2: Let $\alpha = \tan^{-1} (1/5)$ . Then $\tan \alpha = 1/5$ . We want to find $\tan (2\alpha - \pi/4)$ .

Step 3: $\tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \frac{2(1/5)}{1 - (1/25)} = \frac{2/5}{24/25} = \frac{2}{5} \cdot \frac{25}{24} = \frac{5}{12}$ .

Step 4: $\tan (2\alpha - \pi/4) = \frac{\tan 2\alpha - \tan(\pi/4)}{1 + \tan 2\alpha \tan(\pi/4)} = \frac{\frac{5}{12} - 1}{1 + \frac{5}{12}(1)} = \frac{\frac{5-12}{12}}{\frac{12+5}{12}} = \frac{-7}{17}$ .

Step 5: Substitute this into the equation: $17x^2 + 17x \left(-\frac{7}{17}\right) - 10 = 0$ , which simplifies to $17x^2 - 7x - 10 = 0$ .

Step 6: Factor the quadratic: $(17x + 10)(x - 1) = 0$ .

Step 7: The roots are $x = 1$ and $x = -\frac{10}{17}$ .

Step 8: Therefore, a root of the equation is $x = 1$ .

Why other options are wrong:
- Option A: 10/17 is incorrect. It is not a root of the equation.
- Option B: -1 is incorrect. It is not a root of the equation.
- Option C: -7/17 is incorrect. It is not a root of the equation.

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities
- Solving quadratic equations

Memory Aids:
- Remember the formula for $\tan (A - B)$ .

Quick Tips:
- When dealing with trigonometric equations, try to simplify the expression using trigonometric identities.

156 Medium

156 of 581

Application +4 | -1

If

\cos ^{-1} \frac{p}{a}+\cos ^{-1} \frac{q}{b}=\alpha

and

\frac{\mathrm{p}^2}{\mathrm{a}^2}-\frac{2 \mathrm{pq}}{\mathrm{ab}} \cos \alpha+\frac{\mathrm{q}^2}{\mathrm{~b}^2}=\mathrm{f}(\alpha)

, then number of integers in the range of

f(\alpha)

is

View answer and explanation

Correct option: A

2

Option A (1) is correct because by simplifying the integral using trigonometric substitutions, we find that A = 1/2 and B = 1/2, so A+B = 1.

Solution approach

Correct Answer + Core Reason: 1. The value of A+B is calculated by evaluating the integral and comparing it to the given form.

Step-by-Step Solution:

Step 1: Simplify the expression inside the integral.\nLet $x = \cos(2\theta)$ . Then $\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}} = \sqrt{\frac{2\sin^2(\theta)}{2\cos^2(\theta)}} = \tan(\theta)$ .\nSo, $2\tan^{-1}\sqrt{\frac{1-x}{1+x}} = 2\tan^{-1}(\tan(\theta)) = 2\theta$ .

Step 2: Rewrite the integral.
$\int \sin(2\tan^{-1}\sqrt{\frac{1-x}{1+x}}) dx = \int \sin(2\theta) dx$ \nSince $x = \cos(2\theta)$ , $dx = -2\sin(2\theta) d\theta$ .\nSo, the integral becomes:
$\int \sin(2\theta) (-2\sin(2\theta)) d\theta = -2 \int \sin^2(2\theta) d\theta = -2 \int \frac{1 - \cos(4\theta)}{2} d\theta = - \int (1 - \cos(4\theta)) d\theta$
$= -\theta + \frac{1}{4}\sin(4\theta) + C = -\theta + \frac{1}{4}(2\sin(2\theta)\cos(2\theta)) + C = -\theta + \frac{1}{2}\sin(2\theta)\cos(2\theta) + C$ \nSince $x = \cos(2\theta)$ , $\theta = \frac{1}{2}\cos^{-1}(x)$ .\nAlso, $\sin(2\theta) = \sqrt{1 - x^2}$ .\nSo, the integral becomes:
$-\frac{1}{2}\cos^{-1}(x) + \frac{1}{2}\sqrt{1 - x^2}x + C = \frac{1}{2}\sin^{-1}(x) + \frac{1}{2}x\sqrt{1 - x^2} + C - \frac{\pi}{4}$

Step 3: Compare with the given form.\nComparing with $A\sin^{-1}(x) + Bx\sqrt{1 - x^2} + C$ , we have $A = \frac{1}{2}$ and $B = \frac{1}{2}$ .

Step 4: Calculate A + B.\nA + B = $\frac{1}{2} + \frac{1}{2} = 1$

Why other options are wrong:
- Option B (2): Incorrect calculation.
- Option C (3): Incorrect calculation.
- Option D (4): Incorrect calculation.

Key Concepts:
- Trigonometric Substitution
- Integration
- Inverse Trigonometric Functions

Memory Aids:
- Remember the trigonometric identities and substitutions.

Quick Tips:
- Use trigonometric substitution to simplify the integral.

157 Easy

157 of 581

Analyzing +4 | -1

Let

f\left( x \right) = {\left( {\sin \left( {{{\tan }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right)^2} - 1,\,\left| x \right| > 1.\,\,\,\,\mathrm{If }\,\,\frac{{dy}}{{dx}} = \frac{1}{2}\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)\,\,\,\,\mathrm{and}\,\,\,\mathrm{ }y\left( {\sqrt 3 } \right) = \frac{\pi }{6}

then

y\left( { - \sqrt 3 } \right)

is equal to

View answer and explanation

Correct option: B

- \frac{\pi }{6}

Option B (

\frac{{ - \pi }}{6}

) is correct because after simplifying

f(x)

and using the given initial condition, we can find

y(-\sqrt{3})

.

Solution approach

Correct Answer + Core Reason: $- \frac{\pi }{6}$ . The value of $y(-\sqrt{3})$ is found by simplifying $f(x)$ , integrating to find $y(x)$ , using the given condition $y(\sqrt{3}) = \frac{\pi}{6}$ to find the constant of integration, and then evaluating $y(-\sqrt{3})$ .

Step-by-Step Solution:

Step 1: Simplify $f(x)$ .
$f(x) = \left(\sin(\tan^{-1}x) + \sin(\cot^{-1}x)\right)^2 - 1 = \left(\frac{x}{\sqrt{1 + x^2}} + \frac{1}{\sqrt{1 + x^2}}\right)^2 - 1 = \frac{(x + 1)^2}{1 + x^2} - 1 = \frac{x^2 + 2x + 1 - 1 - x^2}{1 + x^2} = \frac{2x}{1 + x^2}$ .

Step 2: Find $y(x)$ using the given differential equation.
$\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} \left(\sin^{-1}(f(x))\right) \Rightarrow 2y = \sin^{-1}(f(x)) + C = \sin^{-1}\left(\frac{2x}{1 + x^2}\right) + C$ .

Step 3: Use the given condition $y(\sqrt{3}) = \frac{\pi}{6}$ to find $C$ .\nSince $|x| > 1$ , $\sin^{-1}\left(\frac{2x}{1 + x^2}\right) = \pi - 2\tan^{-1}x$ for $x > 1$ and $-\pi - 2\tan^{-1}x$ for $x < -1$ .\nFor $x = \sqrt{3}$ , $2y = \pi - 2\tan^{-1}(\sqrt{3}) + C = \pi - 2\left(\frac{\pi}{3}\right) + C = \frac{\pi}{3} + C$ . Since $y(\sqrt{3}) = \frac{\pi}{6}$ , $2\left(\frac{\pi}{6}\right) = \frac{\pi}{3} + C \Rightarrow C = 0$ .

Step 4: Find $y(-\sqrt{3})$ .\nFor $x = -\sqrt{3}$ , $2y = -\pi - 2\tan^{-1}(-\sqrt{3}) + C = -\pi - 2\left(-\frac{\pi}{3}\right) + 0 = -\pi + \frac{2\pi}{3} = -\frac{\pi}{3}$ . Therefore, $y(-\sqrt{3}) = -\frac{\pi}{6}$ .

Why other options are wrong:
- Option A: This is incorrect. It might be obtained from an incorrect calculation of $y(-\sqrt{3})$ .
- Option C: This is incorrect. It might be obtained from an incorrect calculation of $y(-\sqrt{3})$ .
- Option D: This is incorrect. It might be obtained from an incorrect calculation of $y(-\sqrt{3})$ .

Key Concepts:
- Inverse trigonometric functions
- Differentiation
- Integration

Memory Aids:
- $\sin^{-1}\left(\frac{2x}{1 + x^2}\right) = 2\tan^{-1}x$ for $|x| \le 1$ , $\pi - 2\tan^{-1}x$ for $x > 1$ , and $-\pi - 2\tan^{-1}x$ for $x < -1$ .

Quick Tips:
- Remember the properties of inverse trigonometric functions.
- Be careful with signs when solving for $y(-\sqrt{3})$ .

158 Easy

158 of 581

Application +4 | -1

If

\mathrm{\theta } = {\sin ^{ - 1}}x + {\cos ^{ - 1}}x - {\tan ^{ - 1}}x \ge 0,

then the smallest interval in which

\mathrm{\theta }

lies, is given by

View answer and explanation

Correct option: D

\frac{\pi }{4} \le \mathrm{\theta } \le \frac{\pi }{2}

Option D (

\frac{\pi }{4} \le \mathrm{\theta } \le \frac{\pi }{2}

) is correct because, given

\mathrm{\theta } = {\sin ^{ - 1}}x + {\cos ^{ - 1}}x - {\tan ^{ - 1}}x \ge 0

, the smallest interval in which

\mathrm{\theta }

lies is

\frac{\pi }{4} \le \mathrm{\theta } \le \frac{\pi }{2}

.

Solution approach

Correct Answer + Core Reason: $\frac{\pi }{4} \le \mathrm{\theta } \le \frac{\pi }{2}$ . The interval is derived by using the properties of inverse trigonometric functions.

Step-by-Step Solution:

Step 1: Given equation: $\mathrm{\theta } = {\sin ^{ - 1}}x + {\cos ^{ - 1}}x - {\tan ^{ - 1}}x$

Step 2: Use the identity ${{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x = \frac{\pi }{2}$ .
$\mathrm{\theta } = \frac{\pi }{2} - {\tan ^{ - 1}}x$

Step 3: The domain of ${{\sin }^{ - 1}}x$ and ${{\cos }^{ - 1}}x$ is $[-1, 1]$ . Therefore, the domain of $x$ in this problem is $[-1, 1]$ .

Step 4: Find the range of ${{\tan }^{ - 1}}x$ for $x \in [-1, 1]$ .\nWhen $x = -1$ , ${{\tan }^{ - 1}}(-1) = -\frac{\pi }{4}$ .\nWhen $x = 1$ , ${{\tan }^{ - 1}}(1) = \frac{\pi }{4}$ .\nSo, $-\frac{\pi }{4} \le {\tan ^{ - 1}}x \le \frac{\pi }{4}$ .

Step 5: Find the range of $\mathrm{\theta } = \frac{\pi }{2} - {\tan ^{ - 1}}x$ .\nWhen ${{\tan }^{ - 1}}x = -\frac{\pi }{4}$ , $\mathrm{\theta } = \frac{\pi }{2} - \left( { - \frac{\pi }{4}} \right) = \frac{{3\pi }}{4}$ .\nWhen ${{\tan }^{ - 1}}x = \frac{\pi }{4}$ , $\mathrm{\theta } = \frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4}$ .\nHowever, since it's given that $\mathrm{\theta } \ge 0$ , we need to consider the interval where $x$ is in $[-1, 1]

159 Easy

159 of 581

Application +4 | -1

The value of

\tan ^{-1}\left[\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right],|x|<\frac{1}{2}, x \neq 0

, is equal to

View answer and explanation

Correct option: A

\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} \mathrm{x}^{2}

The correct answer is A (

\frac{\pi}{4}+\frac{1}{2} \cos^{-1} \mathrm{x}^{2}

) because it involves using trigonometric substitution and simplification.

Solution approach

Correct Answer + Core Reason: $\frac{{\pi}}{{4}}+\frac{{1}}{{2}} \cos ^{-1} \mathrm{{x}}^{2}$ . The expression is simplified using trigonometric substitution.

Step-by-Step Solution:

Step 1: Let $x^2 = \cos 2\theta$ . Then $\theta = \frac{{1}}{{2}} \cos^{-1} x^2$ .

Step 2: Substitute $x^2 = \cos 2\theta$ into the expression: $\tan^{-1}(\frac{{\sqrt{{1 + x^2}} + \sqrt{{1 - x^2}}}}{{\sqrt{{1 + x^2}} - \sqrt{{1 - x^2}}}}) = \tan^{-1}(\frac{{\sqrt{{1 + \cos 2\theta}} + \sqrt{{1 - \cos 2\theta}}}}{{\sqrt{{1 + \cos 2\theta}} - \sqrt{{1 - \cos 2\theta}}}})$ .

Step 3: Use the identities $1 + \cos 2\theta = 2 \cos^2 \theta$ and $1 - \cos 2\theta = 2 \sin^2 \theta$ : $\tan^{-1}(\frac{{\sqrt{{2 \cos^2 \theta}} + \sqrt{{2 \sin^2 \theta}}}}{{\sqrt{{2 \cos^2 \theta}} - \sqrt{{2 \sin^2 \theta}}}}) = \tan^{-1}(\frac{{\sqrt{{2}} \cos \theta + \sqrt{{2}} \sin \theta}}{{\sqrt{{2}} \cos \theta - \sqrt{{2}} \sin \theta}}) = \tan^{-1}(\frac{{\cos \theta + \sin \theta}}{{\cos \theta - \sin \theta}})$ .

Step 4: Divide both the numerator and denominator by $\cos \theta$ : $\tan^{-1}(\frac{{1 + \tan \theta}}{{1 - \tan \theta}})$ .

Step 5: Use the identity $\tan(A + B) = \frac{{\tan A + \tan B}}{{1 - \tan A \tan B}}$ . Since $\tan \frac{{\pi}}{{4}} = 1$ , we have $\tan^{-1}(\frac{{1 + \tan \theta}}{{1 - \tan \theta}}) = \tan^{-1}(\frac{{\tan \frac{{\pi}}{{4}} + \tan \theta}}{{1 - \tan \frac{{\pi}}{{4}} \tan \theta}}) = \tan^{-1}(\tan(\frac{{\pi}}{{4}} + \theta)) = \frac{{\pi}}{{4}} + \theta$ .

Step 6: Substitute back $\theta = \frac{{1}}{{2}} \cos^{-1} x^2$ : $\frac{{\pi}}{{4}} + \theta = \frac{{\pi}}{{4}} + \frac{{1}}{{2}} \cos^{-1} x^2$ .

Why other options are wrong:
- Option B: Incorrect because the trigonometric substitution was not done correctly.
- Option C: Incorrect because the trigonometric substitution was not done correctly.
- Option D: Incorrect because the trigonometric substitution was not done correctly.

Key Concepts:
- Inverse Trigonometric Functions
- Trigonometric Substitution
- Trigonometric Identities: $1 + \cos 2\theta = 2 \cos^2 \theta$ , $1 - \cos 2\theta = 2 \sin^2 \theta$ , $\tan(A + B) = \frac{{\tan A + \tan B}}{{1 - \tan A \tan B}}$

Memory Aids:
- Remember the trigonometric identities.
- Remember the formula for $\tan(A + B)$ .

Quick Tips:
- Use trigonometric substitution to simplify expressions.
- Use trigonometric identities to simplify expressions.

160 Easy

160 of 581

Application +4 | -1

If

\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}+\ldots.\right)+\cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}+\ldots.\right)=\frac{\pi}{2}

for

0 < |x| < \sqrt{2}

then

x

equal to

View answer and explanation

Correct option: A

1

Option A (1) is correct. The equation simplifies to x = x², leading to x = 1 as the only valid solution within the given range.

Solution approach

Correct Answer + Core Reason: 1. The equation is simplified using geometric series, leading to a quadratic equation with x=1 as the valid solution.

Step-by-Step Solution:

Step 1: We are given $\sin^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}+\ldots.\right) + \cos^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}+\ldots.\right) = \frac{\pi}{2}$ for $0 < |x| < \sqrt{2}$ .

Step 2: Since $\sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2}$ , we must have $x - \frac{x^2}{2} + \frac{x^3}{4} + \ldots = x^2 - \frac{x^4}{2} + \frac{x^6}{4} + \ldots$ .

Step 3: Both series are geometric series. The first series has the first term $x$ and common ratio $-\frac{x}{2}$ , so its sum is $\frac{x}{1 - (-\frac{x}{2})} = \frac{x}{1 + \frac{x}{2}} = \frac{2x}{2+x}$ .

Step 4: The second series has the first term $x^2$ and common ratio $-\frac{x^2}{2}$ , so its sum is $\frac{x^2}{1 - (-\frac{x^2}{2})} = \frac{x^2}{1 + \frac{x^2}{2}} = \frac{2x^2}{2+x^2}$ .

Step 5: Equate the two sums: $\frac{2x}{2+x} = \frac{2x^2}{2+x^2}$ .

Step 6: Simplify the equation: $2x(2+x^2) = 2x^2(2+x)$ , which gives $4x + 2x^3 = 4x^2 + 2x^3$ . This simplifies to $4x = 4x^2$ , so $x = x^2$ .

Step 7: Solve for $x$ : $x^2 - x = 0$ , which gives $x(x-1) = 0$ . The solutions are $x = 0$ and $x = 1$ .

Step 8: Since $0 < |x| < \sqrt{2}$ , $x = 0$ is not a valid solution. Therefore, $x = 1$ .

Why other options are wrong:
- Option B (2): This is incorrect because it does not satisfy the condition $0 < |x| < \sqrt{2}$ .
- Option C (3): This is incorrect because it does not satisfy the condition $0 < |x| < \sqrt{2}$ .
- Option D (4): This is incorrect because it does not satisfy the condition $0 < |x| < \sqrt{2}$ .

Key Concepts:
- Inverse trigonometric functions
- Geometric series
- Solving equations

Memory Aids:
- Remember the formula for the sum of an infinite geometric series: $S = \frac{a}{1-r}$ for $|r| < 1$ .

Quick Tips:
- Recognize geometric series and simplify them before solving the equation.

161 Easy

161 of 581

Application +4 | -1

The principal value of

\sin ^{-1}\left[\sin \left(\frac{2 \pi}{3}\right)\right]

is

View answer and explanation

Correct option: D

None of these

The correct answer is None of these because the principal value of the given expression is

\frac{{\pi}}{{3}}

.

Solution approach

Correct Answer + Core Reason: None of these. The principal value is calculated by finding the equivalent angle within the range of the inverse sine function.

Step-by-Step Solution:

Step 1: Given expression: $\sin ^{-1}\left[\sin \left(\frac{{2 \pi}}{{3}}\right)\right]$

Step 2: We know that $\sin(\theta) = \sin(\pi - \theta)$

Step 3: Therefore, $\sin\left(\frac{{2 \pi}}{{3}}\right) = \sin\left(\pi - \frac{{2 \pi}}{{3}}\right) = \sin\left(\frac{{\pi}}{{3}}\right)$

Step 4: So, $\sin ^{-1}\left[\sin \left(\frac{{2 \pi}}{{3}}\right)\right] = \sin ^{-1}\left[\sin \left(\frac{{\pi}}{{3}}\right)\right]$

Step 5: Since $\frac{{\pi}}{{3}}$ lies in the principal value branch of $\sin^{-1}x$ , which is $[-\frac{{\pi}}{{2}}, \frac{{\pi}}{{2}}]$ , we have: $\sin ^{-1}\left[\sin \left(\frac{{\pi}}{{3}}\right)\right] = \frac{{\pi}}{{3}}$

Why other options are wrong:
- Option A: $-\frac{{2 \pi}}{{3}}$ is incorrect because it's not within the principal value range of $\sin^{-1}x$ .
- Option B: ${}^{\frac{{2 \pi}}{{3}}}$ is incorrect because it's not within the principal value range of $\sin^{-1}x$ .
- Option C: $\frac{{4 \pi}}{{3}}$ is incorrect because it's not within the principal value range of $\sin^{-1}x$ .

Key Concepts:
- Principal value of inverse trigonometric functions
- Properties of sine function

Memory Aids:
- Remember the principal value range of $\sin^{-1}x$ is $[-\frac{{\pi}}{{2}}, \frac{{\pi}}{{2}}]$

162 Medium

162 of 581

Analyzing +4 | -1

If

\tan ^{-1}|a|=\tan ^{-1}\left(a^2+k\right)-\frac{\pi}{4}

holds for integral values of

a \in[-2,2]

. Then number of integral value(s) of

k

is equal to

View answer and explanation

Correct option: A

1

Option A (1) is correct because only one integral value of k satisfies the given equation for the specified range of a.

Solution approach

Correct Answer + Core Reason: 1. Only one integral value of k satisfies the equation.

Step-by-Step Solution:

Step 1: Given equation: $\tan ^{-1}|a|=\tan ^{-1}\left(a^2+k\right)-\frac{\pi}{4}$

Step 2: Rearrange the equation: $\tan ^{-1}\left(a^2+k\right)=\tan ^{-1}|a|+\frac{\pi}{4}$

Step 3: Consider the integral values of $a$ in the range $[-2, 2]$ , i.e., $a = -2, -1, 0, 1, 2$ .

Step 4: Case 1: If $|a|=0$ , then $a=0$ . So, $\tan^{-1}(0+k) = \tan^{-1}(0) + \frac{\pi}{4} = \frac{\pi}{4}$ . This implies $k = \tan(\frac{\pi}{4}) = 1$ .

Step 5: Case 2: If $|a|=1$ , then $a = 1$ or $a = -1$ . So, $\tan^{-1}(1+k) = \tan^{-1}(1) + \frac{\pi}{4} = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$ . This implies $1+k = \tan(\frac{\pi}{2})$ , which is undefined. So, no solution for $k$ in this case.

Step 6: Case 3: If $|a|=2$ , then $a = 2$ or $a = -2$ . So, $\tan^{-1}(4+k) = \tan^{-1}(2) + \frac{\pi}{4}$ .

Step 7: $\tan^{-1}(4+k) = \tan^{-1}(2) + \tan^{-1}(1)$ . Using the formula $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}(\frac{x+y}{1-xy})$ , we have $\tan^{-1}(4+k) = \tan^{-1}(\frac{2+1}{1-2\cdot 1}) = \tan^{-1}(\frac{3}{-1}) = \tan^{-1}(-3)$ .

Step 8: So, $4+k = -3$ , which implies $k = -7$ . However, $\tan^{-1}(a^2+k) = \tan^{-1}(4-7) = \tan^{-1}(-3)$ . Since $a \in [-2,2]$ , $a^2+k$ must be defined. Also, $\tan^{-1}|a| = \tan^{-1}2$ , so $\tan^{-1}2 = \tan^{-1}(-3) - \frac{\pi}{4}$ . This is not possible.

Step 9: The only integral value of $k$ that satisfies the equation is $k=1$ .

Step 10: Final Answer: 1

Why other options are wrong:
- Option B: 2 is incorrect because there is only one integral value of k that satisfies the equation.
- Option C: 3 is incorrect because there is only one integral value of k that satisfies the equation.
- Option D: 0 is incorrect because there is one integral value of k that satisfies the equation.

Key Concepts:
- Inverse trigonometric functions
- Tangent addition formula

Memory Aids:
- Remember the tangent addition formula: $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}(\frac{x+y}{1-xy})$ .

Quick Tips:
- Substitute the given values of $a$ into the equation and solve for $k$ .
- Be careful with the domain and range of inverse trigonometric functions.

163 Easy

163 of 581

Application +4 | -1

If

{\sin ^{ - 1}}x + {\sin ^{ - 1}}\left( {1 - x} \right) = {\cos ^{ - 1}}x

, then

x

belongs to

View answer and explanation

Correct option: C

\left\{ {0,\frac{1}{2}} \right\}

Option C (

\left\{ {0,\frac{1}{2}} \right\}

) is correct because these are the solutions obtained after simplifying the given equation and checking for validity.

Solution approach

Correct Answer + Core Reason: $\left\{ {0,\frac{1}{2}} \right\}$ . The values are obtained by solving the given equation and checking for validity.

Step-by-Step Solution:

Step 1: Given, ${\sin ^{ - 1}}x + {\sin ^{ - 1}}\left( {1 - x} \right) = {\cos ^{ - 1}}x$ .

Step 2: Rewrite the equation as ${\sin ^{ - 1}}\left( {1 - x} \right) = \frac{\pi }{2} - {\sin ^{ - 1}}x - {\sin ^{ - 1}}x = \frac{\pi }{2} - 2{\sin ^{ - 1}}x$ .

Step 3: Use the identity ${\sin ^{ - 1}}a - {\sin ^{ - 1}}b = {\sin ^{ - 1}}\left[ {a\sqrt {1 - {b^2}} - b\sqrt {1 - {a^2}} } \right]$ .
${\sin ^{ - 1}}\left( {1 - x} \right) = {\sin ^{ - 1}}1 - {\sin ^{ - 1}}2x\sqrt {1 - {x^2}} = {\sin ^{ - 1}}\left[ {1\sqrt {1 - 4{x^2}\left( {1 - {x^2}} \right)} - 0} \right]$ .

Step 4: Simplify: $1 - x = \sqrt {1 - 4{x^2} + 4{x^4}} \Rightarrow 1 - x = 1 - 2{x^2} \Rightarrow 2{x^2} - x = 0$ .

Step 5: Solve for x: $x(2x - 1) = 0 \Rightarrow x = 0, \frac{1}{2}$ .

Why other options are wrong:
- Option A: $\left\{ {1,\mathrm{ }0} \right\}$ : This is incorrect because while 0 is a solution, 1 is not a solution to the original equation.
- Option B: $\left\{ { - 1,\mathrm{ }1} \right\}$ : This is incorrect because neither -1 nor 1 are solutions to the original equation.
- Option D: $\left\{ {2,\mathrm{ }0} \right\}$ : This is incorrect because while 0 is a solution, 2 is not a solution to the original equation.

Key Concepts:
- Inverse trigonometric functions
- Solving equations involving inverse trigonometric functions

Memory Aids:
- Remember the identity ${\sin ^{ - 1}}a - {\sin ^{ - 1}}b = {\sin ^{ - 1}}\left[ {a\sqrt {1 - {b^2}} - b\sqrt {1 - {a^2}} } \right]$ .

Quick Tips:
- Always check the validity of the solutions obtained when solving equations involving inverse trigonometric functions.

164 Medium

164 of 581

Analyzing +4 | -1

If

\mathrm{\alpha } = {\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{1}{3}\mathrm{and}\;\mathrm{\beta } = {\cos ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{1}{3},\;

then

View answer and explanation

Correct option: A

\mathrm{\alpha } < \beta

Option A is correct because by calculating alpha and beta and comparing them, it is found that alpha < beta.

Solution approach

Correct Answer + Core Reason: $\mathrm{\alpha } < \beta$ . The relationship between α and β is determined by comparing their values after applying trigonometric identities.

Step-by-Step Solution:

Step 1: Calculate $\alpha = {\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{1}{3} = {\sin ^{ - 1}}(\frac{4}{5}\sqrt {1 - \frac{1}{9}} + \frac{1}{3}\sqrt {1 - \frac{{16}}{{25}}} ) = {\sin ^{ - 1}}(\frac{4}{5}\frac{{2\sqrt 2 }}{3} + \frac{1}{3}\frac{3}{5}) = {\sin ^{ - 1}}(\frac{{8\sqrt 2 + 3}}{{15}})$ .

Step 2: Since $\frac{{8\sqrt 2 + 3}}{{15}} < 1$ , we have $\alpha < \frac{\pi}{2}$ .

Step 3: Calculate $\beta = {\cos ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{1}{3} = \frac{\pi}{2} - {\sin ^{ - 1}}\frac{4}{5} + \frac{\pi}{2} - {\sin ^{ - 1}}\frac{1}{3} = \pi - ({{\sin }^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{1}{3}) = \pi - \alpha$ .

Step 4: Since $\alpha < \frac{\pi}{2}$ , we have $\beta = \pi - \alpha > \alpha$ .

Why other options are wrong:
- Option B: This option is incorrect because α and β are not equal.
- Option C: This option is incorrect because α is less than β.
- Option D: This option is incorrect because one of the other options is correct.

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities

Memory Aids:
- Remember the relationship between sine and cosine inverse functions: ${\cos ^{ - 1}}x = \frac{\pi}{2} - {\sin ^{ - 1}}x$ .

Quick Tips:
- Use trigonometric identities to simplify expressions before comparing them.

165 Easy

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Analyzing +4 | -1

If

\mathrm{\alpha } = {\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{1}{3}\mathrm{and}\;\mathrm{\beta } = {\cos ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{1}{3},\;

then

View answer and explanation

Correct option: A

\mathrm{\alpha } < \beta

Option A (

\alpha < \beta

) is correct because by calculating the values of

\alpha

and

\beta

, we can show that

\alpha

is less than

\beta

.

Solution approach

Correct Answer + Core Reason: $\mathrm{\alpha } < \mathrm{\beta }$ . By calculating the values of $\alpha$ and $\beta$ , we can show that $\alpha$ is less than $\beta$ .

Step-by-Step Solution:

Step 1: Calculate $\alpha$ .
$\mathrm{\alpha } = {\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{1}{3} = {\sin ^{ - 1}}\left( {\frac{4}{5}\sqrt {1 - \frac{1}{9}} + \frac{1}{3}\sqrt {1 - \frac{{16}}{{25}}} } \right) = {\sin ^{ - 1}}\left( {\frac{4}{5} \cdot \frac{{2\sqrt 2 }}{3} + \frac{1}{3} \cdot \frac{3}{5}} \right) = {\sin ^{ - 1}}\left( {\frac{{8\sqrt 2 + 3}}{{15}}} \right)$ \nSince $\frac{{8\sqrt 2 + 3}}{{15}} < 1$ , $\mathrm{\alpha } < \frac{\pi }{2}$

Step 2: Calculate $\beta$ .
$\mathrm{\beta } = {\cos ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{1}{3} = \frac{\pi }{2} - {\sin ^{ - 1}}\frac{4}{5} + \frac{\pi }{2} - {\sin ^{ - 1}}\frac{1}{3} = \pi - \left( {{{\sin }^{ - 1}}\frac{4}{5} + {{\sin }^{ - 1}}\frac{1}{3}} \right) = \pi - \mathrm{\alpha }$

Step 3: Compare $\alpha$ and $\beta$ .\nSince $\mathrm{\alpha } < \frac{\pi }{2}$ , then $\mathrm{\beta } = \pi - \mathrm{\alpha } > \frac{\pi }{2}$ , so $\mathrm{\beta } > \mathrm{\alpha }$

Why other options are wrong:
- Option B: Incorrect. $\alpha$ is not equal to $\beta$ .
- Option C: Incorrect. $\alpha$ is not greater than $\beta$ .
- Option D: The correct answer is among the given options.

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities

Memory Aids:
- Remember that ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$ .

Quick Tips:
- When comparing expressions involving inverse trigonometric functions, try to use trigonometric identities to simplify them.

166 Easy

166 of 581

Application +4 | -1

{\sin ^{ - 1}}\left( {\sin \frac{{2\pi }}{3}} \right) + {\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right) + {\tan ^{ - 1}}\left( {\tan \frac{{3\pi }}{4}} \right)

is equal to:

View answer and explanation

Correct option: A

\frac{{11\pi }}{{12}}

Option A (11π/12) is correct because it involves evaluating inverse trigonometric functions and simplifying the result.

Solution approach

Correct Answer + Core Reason: $\frac{{11\pi }}{{12}}$ . The expression is evaluated by finding the principal values of the inverse trigonometric functions and summing them.

Step-by-Step Solution:

Step 1: Evaluate ${\sin ^{ - 1}}\left( {\sin \frac{{2\pi }}{3}} \right)$ . Since $\frac{{2\pi }}{3}$ is not in the range of ${\sin ^{ - 1}}x$ , we find the equivalent angle in the range $\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$ . ${\sin ^{ - 1}}\left( {\sin \frac{{2\pi }}{3}} \right) = {\sin ^{ - 1}}\left( {\sin \frac{\pi }{3}} \right) = \frac{\pi }{3}$ .

Step 2: Evaluate ${\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right)$ . Since $\frac{{7\pi }}{6}$ is not in the range of ${\cos ^{ - 1}}x$ , we find the equivalent angle in the range $\left[ {0,\pi } \right]$ . ${\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right) = \frac{{5\pi }}{6}$ .

Step 3: Evaluate ${\tan ^{ - 1}}\left( {\tan \frac{{3\pi }}{4}} \right)$ . Since $\frac{{3\pi }}{4}$ is not in the range of ${\tan ^{ - 1}}x$ , we find the equivalent angle in the range $\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$ . ${\tan ^{ - 1}}\left( {\tan \frac{{3\pi }}{4}} \right) = {\tan ^{ - 1}}( - 1) = - \frac{\pi }{4}$ .

Step 4: Sum the results: $\frac{\pi }{3} + \frac{{5\pi }}{6} - \frac{\pi }{4} = \frac{{4\pi + 10\pi - 3\pi }}{{12}} = \frac{{11\pi }}{{12}}$ .

Why other options are wrong:
- Option B: Incorrect evaluation of the inverse trigonometric functions or summation.
- Option C: Incorrect evaluation of the inverse trigonometric functions or summation.
- Option D: Incorrect evaluation of the inverse trigonometric functions or summation.

Key Concepts:
- Inverse trigonometric functions
- Principal values of inverse trigonometric functions

Memory Aids:
- Remember the ranges of inverse trigonometric functions: ${\sin ^{ - 1}}x \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$ , ${\cos ^{ - 1}}x \in \left[ {0,\pi } \right]$ , ${\tan ^{ - 1}}x \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$

Quick Tips:
- Always find the principal values of inverse trigonometric functions before summing them.

167 Easy

167 of 581

Application +4 | -1

The value of

\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)

is equal to

View answer and explanation

Correct option: B

105^{\circ}

Option B (

105^{\circ}

) is correct because

\sin^{-1}(\frac{\sqrt{3}}{2}) + \sin^{-1}(\frac{1}{\sqrt{2}})

equals

\frac{\pi}{3} + \frac{\pi}{4}

, which simplifies to

\frac{7\pi}{12}

or

105^{\circ}

.

Solution approach

Correct Answer + Core Reason: $105^{\circ}$ . The sum of the inverse sines is calculated by converting to radians and adding.

Step-by-Step Solution:

Step 1: Identify the values of the inverse sine functions.
$\sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$ or $60^{\circ}$
$\sin^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$ or $45^{\circ}$

Step 2: Add the two angles in radians.
$\frac{\pi}{3} + \frac{\pi}{4} = \frac{4\pi + 3\pi}{12} = \frac{7\pi}{12}$

Step 3: Convert the result to degrees.
$\frac{7\pi}{12} \times \frac{180^{\circ}}{\pi} = 7 \times 15^{\circ} = 105^{\circ}$

Final Answer: $105^{\circ}$

Why other options are wrong:
- **Option A ( $75^{\circ}$ ): This is incorrect because it results from subtracting the angles or using incorrect values for the inverse sines.
- Option C ( $\frac{5 \pi}{12}$ ): This is incorrect because it results from an incorrect addition of the fractions or using incorrect values for the inverse sines.
- Option D ( $\frac{3 \pi}{5}$ ): This is incorrect because it results from using incorrect values for the inverse sines and performing an incorrect calculation.

Key Concepts:
- Inverse trigonometric functions
- Radian to degree conversion

Memory Aids:
- Remember the common values of sine and their corresponding angles in radians and degrees.

Quick Tips:**
- Convert inverse trigonometric functions to radians first, then add, and finally convert back to degrees if needed.

168 Easy

168 of 581

Application +4 | -1

If

- 1 \le x \le 0

, then

{\cos ^{ - 1}}\left( {2{x^2} - 1} \right)

equals

View answer and explanation

Correct option: C

2\;\pi - 2{\cos ^{ - 1}}x

The correct answer is C (

2\;\pi - 2{\cos ^{ - 1}}x

) because considering the given range of x,

{\cos ^{ - 1}}(2x^2 - 1)

simplifies to

2\pi - 2{\cos ^{ - 1}}x

.

Solution approach

Correct Answer + Core Reason: $2\;\pi - 2{\cos ^{ - 1}}x$ . The expression simplifies to this form considering the given range of x.

Step-by-Step Solution:

Step 1: Let ${\cos ^{ - 1}}x = \theta$ . Then $x = \cos \theta$ .

Step 2: Given $- 1 \le x \le 0$ , which implies $- 1 \le \cos \theta \le 0$ . This means $\frac{\pi }{2} \le \theta \le \pi$ .

Step 3: Now, consider ${\cos ^{ - 1}}(2{x^2} - 1) = {\cos ^{ - 1}}(2{\cos ^2}\theta - 1) = {\cos ^{ - 1}}(\cos 2\theta)$ .

Step 4: Since $\frac{\pi }{2} \le \theta \le \pi$ , then $\pi \le 2\theta \le 2\pi$ . To bring $2\theta$ within the range of ${\cos ^{ - 1}}$ , which is $[0, \pi]$ , we use $2\pi - 2\theta$ .

Step 5: Therefore, ${\cos ^{ - 1}}(\cos 2\theta) = 2\pi - 2\theta$ .

Step 6: Substitute back $\theta = {\cos ^{ - 1}}x$ : $2\pi - 2{\cos ^{ - 1}}x$ .

Why other options are wrong:
- Option A: $2{\cos ^{ - 1}}x$ : This is incorrect because it doesn't account for the range of x.
- Option B: $\pi - 2{\cos ^{ - 1}}x$ : This is incorrect because it doesn't account for the range of x.
- Option D: $- 2{\cos ^{ - 1}}x$ : This is incorrect because it doesn't account for the range of x and also has the wrong sign.

Key Concepts:
- Inverse Trigonometric Functions
- Trigonometric Identities
- Range of Inverse Cosine Function

Memory Aids:
- Remember the identity $\cos 2\theta = 2{\cos ^2}\theta - 1$ .
- Keep in mind the range of ${\cos ^{ - 1}}x$ is $[0, \pi]$ .

Quick Tips:
- Always consider the range of the variable when dealing with inverse trigonometric functions.
- Use trigonometric identities to simplify the expression.

169 Medium

169 of 581

Application +4 | -1

The principal value of

\sin ^{-1}(-\sqrt{3} / 2)+\cos ^{-1} \cos (7 \pi / 6)

is

View answer and explanation

Correct option: B

\pi / 2

Option B (

\pi / 2

) is correct because

\sin^{-1}(-\sqrt{3} / 2) = -\pi / 3

and

\cos^{-1}(\cos(7\pi / 6)) = 5\pi / 6

, and their sum is

\pi / 2

.

Solution approach

Correct Answer + Core Reason: $\pi / 2$ . The principal value is calculated by evaluating each inverse trigonometric function separately and then summing the results.

Step-by-Step Solution:

Step 1: Evaluate $\sin^{-1}(-\sqrt{3} / 2)$ . Since $\sin(-\pi / 3) = -\sqrt{3} / 2$ , and $-\pi / 3$ is in the range of $\sin^{-1}(x)$ , we have $\sin^{-1}(-\sqrt{3} / 2) = -\pi / 3$ .

Step 2: Evaluate $\cos^{-1}(\cos(7\pi / 6))$ . Since $7\pi / 6$ is not in the range of $\cos^{-1}(x)$ , which is $[0, \pi]$

170 Easy

170 of 581

Understanding +4 | -1

The value of

{\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) - {\sin ^{ - 1}}\left( {\frac{1}{2}} \right)

is

View answer and explanation

Correct option: D

30^\circ

Option D (

30^\circ

) is correct because

{\sin ^{ - 1}}(\frac{{\sqrt 3 }}{2}) = 60^\circ

and

{\sin ^{ - 1}}(\frac{1}{2}) = 30^\circ

, so the difference is

60^\circ - 30^\circ = 30^\circ

.

Solution approach

Correct Answer + Core Reason: $30^\circ$ . The difference between the two inverse sine values is 30 degrees.

Step-by-Step Solution:

Step 1: Evaluate ${\sin ^{ - 1}}(\frac{{\sqrt 3 }}{2})$ . We know that $\sin(60^\circ) = \frac{{\sqrt 3 }}{2}$ , so ${\sin ^{ - 1}}(\frac{{\sqrt 3 }}{2}) = 60^\circ$ .

Step 2: Evaluate ${\sin ^{ - 1}}(\frac{1}{2})$ . We know that $\sin(30^\circ) = \frac{1}{2}$ , so ${\sin ^{ - 1}}(\frac{1}{2}) = 30^\circ$ .

Step 3: Calculate the difference: ${\sin ^{ - 1}}(\frac{{\sqrt 3 }}{2}) - {\sin ^{ - 1}}(\frac{1}{2}) = 60^\circ - 30^\circ = 30^\circ$ .

Why other options are wrong:
- Option A: This is incorrect because the difference is not 45 degrees.
- Option B: This is incorrect because the difference is not 90 degrees.
- Option C: This is incorrect because the difference is not 15 degrees.

Key Concepts:
- Inverse trigonometric functions
- Evaluation of trigonometric functions

Memory Aids:
- Remember the common values of sine for standard angles.

Quick Tips:
- Know the values of inverse trigonometric functions for common angles.

171 Easy

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Application +4 | -1

If

\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=-\frac{3 \pi}{2}

then the value of

x y+y z+z x

is

View answer and explanation

Correct option: A

3

The correct answer is A (3) because when

\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = -\frac{3\pi}{2}

, it implies x = y = z = -1, and therefore xy + yz + zx = 3.

Solution approach

Correct Answer + Core Reason: 3. Since the maximum value of $\sin^{-1}x$ is $\frac{\pi}{2}$ , the only way for the sum to be $-\frac{3\pi}{2}$ is if each term is $-\frac{\pi}{2}$ .

Step-by-Step Solution:

Step 1: Analyze the given equation: $\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = -\frac{3\pi}{2}$ .

Step 2: Determine the values of x, y, and z: Since the range of $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ , the only way for the sum to be $-\frac{3\pi}{2}$ is if $\sin^{-1}x = \sin^{-1}y = \sin^{-1}z = -\frac{\pi}{2}$ .

Step 3: Solve for x, y, and z: $x = \sin(-\frac{\pi}{2}) = -1$ , $y = \sin(-\frac{\pi}{2}) = -1$ , $z = \sin(-\frac{\pi}{2}) = -1$ .

Step 4: Calculate xy + yz + zx: $xy + yz + zx = (-1)(-1) + (-1)(-1) + (-1)(-1) = 1 + 1 + 1 = 3$ .

Why other options are wrong:
- Option B: This is incorrect because the calculation is wrong.
- Option C: This is incorrect because the calculation is wrong.
- Option D: This is incorrect because the calculation is wrong.

Key Concepts:
- Range of arcsine function
- Solving trigonometric equations

Memory Aids:
- Remember that the range of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ .

Quick Tips:
- When dealing with sums of inverse trigonometric functions, consider the range of each function.

172 Easy

172 of 581

Analyzing +4 | -1

If

{\cos ^{ - 1}}x = \alpha ,\;(0 < x < 1)

and

{\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) + {\sec ^{ - 1}}\left( {\frac{1}{{2{x^2} - 1}}} \right) = \frac{{2\pi }}{3}

, then

{\tan ^{ - 1}}\left( {2x} \right)

equals

View answer and explanation

Correct option: C

\pi /3

Option C is correct because it solves for alpha using the given equation and then finds tan^-1(2x) by substituting the value of x.

Solution approach

Correct Answer + Core Reason: $\pi /3$ . The value of $\tan^{-1}(2x)$ is found by solving for x and substituting it into the expression.

Step-by-Step Solution:

Step 1: Given ${\cos ^{ - 1}}x = \alpha$ , so $x = \cos \alpha$ .

Step 2: Substitute $x = \cos \alpha$ into the equation ${\sin ^{ - 1}}(2x\sqrt{1 - x^2}) + {\sec ^{ - 1}}(\frac{1}{{2x^2 - 1}}) = \frac{{2\pi}}{3}$ .

Step 3: ${\sin ^{ - 1}}(2\cos \alpha \sqrt{1 - {{\cos }^2}\alpha }) + {\sec ^{ - 1}}(\frac{1}{{2{{\cos }^2}\alpha - 1}}) = \frac{{2\pi}}{3}$ simplifies to ${\sin ^{ - 1}}(\sin 2\alpha) + {\sec ^{ - 1}}(\sec 2\alpha) = \frac{{2\pi}}{3}$ .

Step 4: $2\alpha + 2\alpha = \frac{{2\pi}}{3}$ , so $4\alpha = \frac{{2\pi}}{3}$ , which gives $\alpha = \frac{\pi}{6}$ .

Step 5: $x = \cos \frac{\pi}{6} = \frac{{\sqrt 3 }}{2}$ , so $2x = \sqrt 3$ .

Step 6: Therefore, ${\tan ^{ - 1}}(2x) = {\tan ^{ - 1}}(\sqrt 3) = \frac{\pi}{3}$ .

Why other options are wrong:
- Option A: This option is incorrect because it leads to a different value for x and thus a different value for $\tan^{-1}(2x)$ .
- Option B: This option is incorrect because it leads to a different value for x and thus a different value for $\tan^{-1}(2x)$ .
- Option D: This option is incorrect because it leads to a different value for x and thus a different value for $\tan^{-1}(2x)$ .

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities

Memory Aids:
- Remember the values of trigonometric functions for standard angles.

Quick Tips:
- Simplify the equation using trigonometric identities before solving for the variable.

173 Easy

173 of 581

Application +4 | -1

The value of

\sin \left( {4{{\tan }^{ - 1}}\frac{1}{3}} \right) - \cos \left( {2{{\tan }^{ - 1}}\frac{1}{7}} \right)

is

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the value of the expression simplifies to 0, which is not among the given options.

Solution approach

Correct Answer + Core Reason: None of these. The value of the expression simplifies to 0, which is not in the options.

Step-by-Step Solution:

Step 1: Evaluate $\sin \left( {4{{\tan }^{ - 1}}\frac{1}{3}} \right)$ . Let $\theta = {\tan ^{ - 1}}\frac{1}{3}$ . Then, $\tan \theta = \frac{1}{3}$ .

Step 2: $\sin (4\theta ) = 2\sin (2\theta )\cos (2\theta )$ .

Step 3: $\sin (2\theta ) = \frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \frac{{2\left( {\frac{1}{3}} \right)}}{{1 + {{\left( {\frac{1}{3}} \right)}^2}}} = \frac{{\frac{2}{3}}}{{1 + \frac{1}{9}}} = \frac{{\frac{2}{3}}}{{\frac{{10}}{9}}} = \frac{2}{3} \times \frac{9}{{10}} = \frac{3}{5}$ .

Step 4: $\cos (2\theta ) = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \frac{{1 - {{\left( {\frac{1}{3}} \right)}^2}}}{{1 + {{\left( {\frac{1}{3}} \right)}^2}}} = \frac{{1 - \frac{1}{9}}}{{1 + \frac{1}{9}}} = \frac{{\frac{8}{9}}}{{\frac{{10}}{9}}} = \frac{8}{{10}} = \frac{4}{5}$ .

Step 5: $\sin (4\theta ) = 2\sin (2\theta )\cos (2\theta ) = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{{24}}{{25}}$ .

Step 6: Evaluate $\cos \left( {2{{\tan }^{ - 1}}\frac{1}{7}} \right)$ . Let $\phi = {\tan ^{ - 1}}\frac{1}{7}$ . Then, $\tan \phi = \frac{1}{7}$ .

Step 7: $\cos (2\phi ) = \frac{{1 - {{\tan }^2}\phi }}{{1 + {{\tan }^2}\phi }} = \frac{{1 - {{\left( {\frac{1}{7}} \right)}^2}}}{{1 + {{\left( {\frac{1}{7}} \right)}^2}}} = \frac{{1 - \frac{1}{{49}}}}{{1 + \frac{1}{{49}}}} = \frac{{\frac{{48}}{{49}}}}{{\frac{{50}}{{49}}}} = \frac{{48}}{{50}} = \frac{{24}}{{25}}$ .

Step 8: $\sin \left( {4{{\tan }^{ - 1}}\frac{1}{3}} \right) - \cos \left( {2{{\tan }^{ - 1}}\frac{1}{7}} \right) = \frac{{24}}{{25}} - \frac{{24}}{{25}} = 0$ .

Why other options are wrong:
- Option A: $\frac{3}{7}$ is incorrect because it is not the correct value of the expression.
- Option B: $\frac{7}{8}$ is incorrect because it is not the correct value of the expression.
- Option C: $\frac{8}{{21}}$ is incorrect because it is not the correct value of the expression.

Key Concepts:
- Inverse Trigonometric Functions
- Trigonometric Identities (Double Angle Formula)

Memory Aids:
- Remember the double angle formulas for sine and cosine in terms of tangent.

Quick Tips:
- Use trigonometric identities to simplify expressions before evaluating inverse trigonometric functions.
- Be careful with algebraic manipulations and simplifications.

174 Medium

174 of 581

Analyzing +4 | -1

Statement - I If

x=\sin ^{-1}(\sin 10)

and

y=\cos ^{-1}(\cos 10)

then

y-x=\pi

Statement - II The no. of solutions of the equation

\sin ^{-1}\left[x^2+\frac{1}{3}\right]+\cos ^{-1}\left[x^2-\frac{2}{3}\right]=x^2, x \in[-1,1]

and [.] denoted the greatest integer less than and equal to

\mathrm{x}

is 0

View answer and explanation

Correct option: C

Both statements are true

Option C (Both statements are true) is correct because both the simplification of inverse trigonometric functions and the analysis of the equation involving greatest integer functions lead to valid conclusions.

Solution approach

Correct Answer + Core Reason: Both statements are true. Statement I simplifies correctly using properties of inverse trigonometric functions, and Statement II is proven to have no solutions.

Step-by-Step Solution:

Statement I:
Step 1: Simplify $x = \sin^{-1}(\sin 10)$ .\nSince $3\pi < 10 < \frac{7\pi}{2}$ , we have $10 = 3\pi - (3\pi - 10)$ .\nThus, $x = \sin^{-1}(\sin(3\pi - 10)) = 3\pi - 10$ .

Step 2: Simplify $y = \cos^{-1}(\cos 10)$ .\nSince $3\pi < 10 < \frac{7\pi}{2}$ , we have $10 = 4\pi - (4\pi - 10)$ .\nThus, $y = \cos^{-1}(\cos(4\pi - 10)) = 4\pi - 10$ .

Step 3: Calculate $y - x$ .
$y - x = (4\pi - 10) - (3\pi - 10) = \pi$ .\nTherefore, Statement I is true.

Statement II:
Step 1: Analyze the equation $\sin^{-1}[x^2 + \frac{1}{3}] + \cos^{-1}[x^2 - \frac{2}{3}] = x^2$ .\nRewrite the equation as $\sin^{-1}[x^2 + \frac{1}{3}] + \cos^{-1}([x^2 + \frac{1}{3}] - 1) = x^2$ .

Step 2: Determine the possible values of $[x^2 + \frac{1}{3}]$ .\nSince $x \in [-1, 1]$ , we have $0 \le x^2 \le 1$ , so $\frac{1}{3} \le x^2 + \frac{1}{3} \le \frac{4}{3}$ .\nTherefore, $[x^2 + \frac{1}{3}]$ can be 0 or 1.

Step 3: Check the case when $[x^2 + \frac{1}{3}] = 0$ .
$\sin^{-1}(0) + \cos^{-1}(-1) = x^2 \Rightarrow 0 + \pi = x^2 \Rightarrow x^2 = \pi$ , which is not possible since $0 \le x^2 \le 1$ .

Step 4: Check the case when $[x^2 + \frac{1}{3}] = 1$ .
$\sin^{-1}(1) + \cos^{-1}(0) = x^2 \Rightarrow \frac{\pi}{2} + \frac{\pi}{2} = x^2 \Rightarrow x^2 = \pi$ , which is not possible since $0 \le x^2 \le 1$ .

Step 5: Conclude that the equation has no solutions.\nTherefore, Statement II is true.

Why other options are wrong:
- Option A (Statement - I is true): Incorrect because Statement II is also true.
- Option B (Statement - II is true): Incorrect because Statement I is also true.
- Option D (None): Incorrect because both statements are true.

Key Concepts:
- Inverse trigonometric functions
- Greatest integer function

Memory Aids:
- Remember the range of inverse sine and cosine functions.
- Understand the properties of the greatest integer function.

Quick Tips:
- Simplify inverse trigonometric expressions carefully.
- Consider the possible values of the greatest integer function.

175 Medium

175 of 581

Application +4 | -1

If

{\sec ^{ - 1}}\sqrt {1 + {x^2}} + \mathrm{cose}{\mathrm{c}^{ - 1}}\frac{{\sqrt {1 + {y^2}} }}{y} + {\cot ^{ - 1}}\frac{1}{z} = \pi

, then

x + y + z

is equal to

View answer and explanation

Correct option: A

xyz

Option A (

xyz

) is correct because the given equation can be transformed into

{\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi

, which implies

x + y + z = xyz

.

Solution approach

Correct Answer + Core Reason: $xyz$ . The given equation simplifies to $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi$ , which leads to $x+y+z = xyz$ .

Step-by-Step Solution:

Step 1: Rewrite the given equation using inverse trigonometric identities.

Step 2: ${\sec ^{ - 1}}\sqrt {1 + {x^2}} = {\tan ^{ - 1}}x$ , $\mathrm{cose}{\mathrm{c}^{ - 1}}\frac{{\sqrt {1 + {\mathrm{y}^2}} }}{{\mathrm{y}}} = {\tan ^{ - 1}}y$ , and ${\cot ^{ - 1}}\frac{1}{z} = {\tan ^{ - 1}}z$ .

Step 3: Substitute these identities into the given equation: ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi$ .

Step 4: Apply the formula for the sum of three arctangents: ${\tan ^{ - 1}}\left( {\frac{{x + y + z - xyz}}{{1 - xy - yz - zx}}} \right) = \pi$ .

Step 5: Take the tangent of both sides: $\frac{{x + y + z - xyz}}{{1 - xy - yz - zx}} = \tan(\pi) = 0$ .

Step 6: Therefore, $x + y + z - xyz = 0$ , which implies $x + y + z = xyz$ .

Why other options are wrong:
- Option B: $2xyz$ is incorrect because the correct relationship derived from the given equation is $x + y + z = xyz$ , not $x + y + z = 2xyz$ .
- Option C: $xy{z^2}$ is incorrect because the correct relationship derived from the given equation is $x + y + z = xyz$ , not $x + y + z = xy{z^2}$ .
- Option D: ${x^2}yz$ is incorrect because the correct relationship derived from the given equation is $x + y + z = xyz$ , not $x + y + z = {x^2}yz$ .

Key Concepts:
- Inverse trigonometric functions
- ${\sec ^{ - 1}}\sqrt {1 + {x^2}} = {\tan ^{ - 1}}x$
- $\mathrm{cose}{\mathrm{c}^{ - 1}}\frac{{\sqrt {1 + {\mathrm{y}^2}} }}{{\mathrm{y}}} = {\tan ^{ - 1}}y$
- ${\cot ^{ - 1}}\frac{1}{z} = {\tan ^{ - 1}}z$
- ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = {\tan ^{ - 1}}\left( {\frac{{x + y + z - xyz}}{{1 - xy - yz - zx}}} \right)$

Memory Aids:
- Remember the identities for inverse trigonometric functions.
- Remember the formula for the sum of three arctangents.

Quick Tips:
- Convert all inverse trigonometric functions to arctangent.
- Use the formula for the sum of three arctangents to simplify the equation.

176 Medium

176 of 581

Application +4 | -1

If

{\sin ^{ - 1}}x - {\cos ^{ - 1}}x = \frac{\pi }{6},

then

x

is

View answer and explanation

Correct option: B

\frac{{\sqrt 3 }}{2}

Option B (

\frac{{\sqrt 3 }}{2}

) is correct because solving the equation using trigonometric identities leads to this value for x.

Solution approach

Correct Answer + Core Reason: $\frac{{\sqrt 3 }}{2}$ . Solving the equation using trigonometric identities leads to this value for x.

Step-by-Step Solution:

Step 1: Given ${\sin ^{ - 1}}x - {\cos ^{ - 1}}x = \frac{\pi }{6}$ .

Step 2: Use the identity ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$ to rewrite ${\sin ^{ - 1}}x$ as $\frac{\pi }{2} - {\cos ^{ - 1}}x$ .

Step 3: Substitute into the given equation: $(\frac{\pi }{2} - {\cos ^{ - 1}}x) - {\cos ^{ - 1}}x = \frac{\pi }{6}$ .

Step 4: Simplify: $\frac{\pi }{2} - 2{\cos ^{ - 1}}x = \frac{\pi }{6} \Rightarrow 2{\cos ^{ - 1}}x = \frac{\pi }{2} - \frac{\pi }{6} = \frac{{3\pi - \pi }}{6} = \frac{{2\pi }}{6} = \frac{\pi }{3}$ .

Step 5: Solve for ${\cos ^{ - 1}}x$ : ${\cos ^{ - 1}}x = \frac{\pi }{6}$ .

Step 6: Solve for $x$ : $x = \cos \frac{\pi }{6} = \frac{{\sqrt 3 }}{2}$ .

Why other options are wrong:
- Option A: $\frac{1}{2}$ is incorrect because substituting $x=\frac{1}{2}$ into the original equation does not satisfy the equation.
- Option C: $- \frac{1}{2}$ is incorrect because substituting $x=-\frac{1}{2}$ into the original equation does not satisfy the equation.
- Option D: None of these is incorrect because the correct answer is $\frac{{\sqrt 3 }}{2}$ .

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities

Memory Aids:
- Remember that ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$ .
- Know the values of trigonometric functions at common angles.

Quick Tips:
- Use the identity ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$ to simplify equations involving both ${\sin ^{ - 1}}x$ and ${\cos ^{ - 1}}x$ .
- Know the values of trigonometric functions for common angles.

177 Medium

177 of 581

Application +4 | -1

If

\tan ^{-1}(a / x)+\tan ^{-1}(b / x)=\pi / 2

, then

x

is equal to

View answer and explanation

Correct option: D

\sqrt{a b}

Option D,

\sqrt{ab}

, is correct because by applying the tangent addition formula and using the given condition, we can derive that

x = \sqrt{ab}

.

Solution approach

Correct Answer + Core Reason: $\sqrt{a b}$ . The value of x is found by using the tangent addition formula and the given condition.

Step-by-Step Solution:

Step 1: Given $\tan^{-1}(\frac{{a}}{{x}}) + \tan^{-1}(\frac{{b}}{{x}}) = \frac{{\pi}}{{2}}$ .

Step 2: Apply the tangent addition formula: $\tan^{-1}(\frac{{\frac{{a}}{{x}} + \frac{{b}}{{x}}}}{{1 - \frac{{a}}{{x}} \cdot \frac{{b}}{{x}}}}) = \frac{{\pi}}{{2}}$ .

Step 3: This implies $\frac{{\frac{{a + b}}{{x}}}}{{1 - \frac{{ab}}{{x^2}}}} = \tan(\frac{{\pi}}{{2}})$ , which is undefined. This means the denominator must be zero.

Step 4: Therefore, $1 - \frac{{ab}}{{x^2}} = 0$ , which implies $x^2 = ab$ .

Step 5: Taking the square root of both sides, we get $x = \sqrt{{ab}}$ .

Why other options are wrong:
- Option A: $ab$ is incorrect because it does not satisfy the equation when substituted back into the original equation.
- Option B: $a/b$ is incorrect because it does not satisfy the equation when substituted back into the original equation.
- Option C: $b/a$ is incorrect because it does not satisfy the equation when substituted back into the original equation.

Key Concepts:
- Inverse trigonometric functions
- Tangent addition formula: $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}(\frac{{x + y}}{{1 - xy}})$

Memory Aids:
- Remember the tangent addition formula.

Quick Tips:
- When dealing with inverse trigonometric functions summing to $\pi/2$ , the denominator of the tangent addition formula will be zero.

178 Easy

178 of 581

Application +4 | -1

If

{\cos ^{ - 1}}\sqrt p + {\cos ^{ - 1}}\sqrt {1 - p} + {\cos ^{ - 1}}\sqrt {1 - q} = \frac{{3\pi }}{4},

then the value of

q

is

View answer and explanation

Correct option: D

\frac{1}{2}

Option D (

\frac{1}{2}

) is correct because by using the given equation and trigonometric identities, we can derive that

q = \frac{1}{2}

.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2}$ . By using the given equation and trigonometric identities, we can derive that $q = \frac{1}{2}$ .

Step-by-Step Solution:

Step 1: Let $\mathrm{\alpha } = {\cos ^{ - 1}}\sqrt p ,\mathrm{\beta } = {\cos ^{ - 1}}\sqrt {1 - p} ,\mathrm{\gamma } = {\cos ^{ - 1}}\sqrt {1 - q}$ .\nThen, $\cos \mathrm{\alpha } = \sqrt p ,\cos \mathrm{\beta } = \sqrt {1 - p} ,\cos \mathrm{\gamma } = \sqrt {1 - q}$ .

Step 2: Find $\sin{\alpha}$ , $\sin{\beta}$ , and $\sin{\gamma}$ .
$\sin \mathrm{\alpha } = \sqrt {1 - p} ,\sin \mathrm{\beta } = \sqrt p ,\sin \mathrm{\gamma } = \sqrt q$

Step 3: Use the given equation to relate $\alpha$ , $\beta$ , and $\gamma$ .
$\mathrm{\alpha } + \mathrm{\beta } + \mathrm{\gamma } = \frac{{3\pi }}{4} \Rightarrow \mathrm{\alpha } + \mathrm{\beta } = \frac{{3\pi }}{4} - \mathrm{\gamma }$

Step 4: Take the cosine of both sides.
$\cos \left( {\mathrm{\alpha } + \mathrm{\beta }} \right) = \cos \left( {\frac{{3\pi }}{4} - \mathrm{\gamma }} \right)$

Step 5: Use the cosine addition formula.
$\cos \mathrm{\alpha }\cos \mathrm{\beta } - \sin \mathrm{\alpha }\sin \mathrm{\beta } = \cos \frac{{3\pi }}{4}\cos \mathrm{\gamma } + \sin \frac{{3\pi }}{4}\sin \mathrm{\gamma }$
$\sqrt p \sqrt {1 - p} - \sqrt {1 - p} \sqrt p = - \frac{1}{{\sqrt 2 }}\sqrt {1 - q} + \frac{1}{{\sqrt 2 }}\sqrt q$

Step 6: Simplify the equation.
$0 = - \frac{1}{{\sqrt 2 }}\sqrt {1 - q} + \frac{1}{{\sqrt 2 }}\sqrt q \Rightarrow \sqrt {1 - q} = \sqrt q \Rightarrow 1 - q = q \Rightarrow 2q = 1 \Rightarrow q = \frac{1}{2}$

Why other options are wrong:
- Option A (1): This is incorrect because it doesn't satisfy the original equation.
- **Option B ( $\frac{1}{{\sqrt 2 }}$ ): This is incorrect because it doesn't satisfy the original equation.
- Option C ( $\frac{1}{3}$ ): This is incorrect because it doesn't satisfy the original equation.

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities

Memory Aids:**
- Remember the cosine addition formula: $\cos(A+B) = \cos A \cos B - \sin A \sin B$

Quick Tips:
- When dealing with inverse trigonometric functions and given equations, try to express the variables in terms of trigonometric functions and use the given equation to relate them.

179 Medium

179 of 581

Application +4 | -1

If

\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}

, then

x

is equal to

View answer and explanation

Correct option: C

\frac{2}{3},-\frac{2}{3}

Option C (

\frac{2}{3}, -\frac{2}{3}

) is correct because solving the trigonometric equation

\cos(2\sin^{-1}x) = \frac{1}{9}

yields both positive and negative solutions for x.

Solution approach

Correct Answer + Core Reason: $\frac{2}{3},-\frac{2}{3}$ . Solving the equation $\cos(2\sin^{-1} x) = \frac{1}{9}$ gives both positive and negative solutions.

Step-by-Step Solution:

Step 1: Let $\sin^{-1} x = \theta$ . Then $x = \sin \theta$ .

Step 2: Substitute into the given equation: $\cos(2\theta) = \frac{1}{9}$ .

Step 3: Use the identity $\cos(2\theta) = 1 - 2\sin^2 \theta$ .
$1 - 2\sin^2 \theta = \frac{1}{9}$ .

Step 4: Substitute $x = \sin \theta$ back into the equation.
$1 - 2x^2 = \frac{1}{9}$ .

Step 5: Solve for $x^2$ .
$2x^2 = 1 - \frac{1}{9} = \frac{8}{9}$ .
$x^2 = \frac{4}{9}$ .

Step 6: Solve for $x$ .
$x = \pm \frac{2}{3}$ .

Why other options are wrong:
- **Option A (Only $\frac{2}{3}$ ): This is incorrect because it only considers the positive root and neglects the negative root.
- Option B (Only $-\frac{2}{3}$ ): This is incorrect because it only considers the negative root and neglects the positive root.
- Option D (Neither $\frac{2}{3}$ nor $-\frac{2}{3}$ ):** This is incorrect because both $\frac{2}{3}$ and $-\frac{2}{3}$ are valid solutions.

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities
- Solving trigonometric equations

Memory Aids:
- Remember the identity $\cos(2\theta) = 1 - 2\sin^2 \theta$ .

Quick Tips:
- Always consider both positive and negative roots when solving for a variable squared.

180 Easy

180 of 581

Application +4 | -1

A solution of the equation

{\tan ^{ - 1}}\left( {1 + x} \right) + {\tan ^{ - 1}}\left( {1 - x} \right) = \frac{\pi }{2},

is

View answer and explanation

Correct option: C

x = 0

Option C is correct because solving the equation

{\tan ^{ - 1}}\left( {1 + x} \right) + {\tan ^{ - 1}}\left( {1 - x} \right) = \frac{\pi }{2}

using trigonometric identities leads to

x = 0

.

Solution approach

Correct Answer + Core Reason: $x = 0$ . Solving the given equation using trigonometric identities leads to this value.

Step-by-Step Solution:

Step 1: Use the identity ${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\frac{{a + b}}{{1 - ab}}} \right)$ .
${\tan ^{ - 1}}\left( {1 + x} \right) + {\tan ^{ - 1}}\left( {1 - x} \right) = {\tan ^{ - 1}}\left( {\frac{{(1 + x) + (1 - x)}}{{1 - (1 + x)(1 - x)}}} \right) = \frac{\pi }{2}$

Step 2: Simplify the expression inside the inverse tangent.
${\tan ^{ - 1}}\left( {\frac{2}{{1 - (1 - {x^2})}}} \right) = {\tan ^{ - 1}}\left( {\frac{2}{{{x^2}}}} \right) = \frac{\pi }{2}$

Step 3: Take the tangent of both sides.
$\frac{2}{{{x^2}}} = \tan \frac{\pi }{2} = \infty$

Step 4: For the fraction to be infinite, the denominator must be zero.
${x^2} = 0 \Rightarrow x = 0$

Why other options are wrong:
- Option A: This option is incorrect because $x = 1$ does not satisfy the given equation.
- Option B: This option is incorrect because $x = -1$ does not satisfy the given equation.
- Option D: This option is incorrect because $x = \pi$ does not satisfy the given equation.

Key Concepts:
- Inverse Trigonometric Functions
- Tangent Addition Formula

Memory Aids:
- Remember the tangent addition formula: ${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\frac{{a + b}}{{1 - ab}}} \right)$

Quick Tips:
- When solving equations involving inverse trigonometric functions, try to use trigonometric identities to simplify the equation.

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