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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Medium

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121 Medium
121 of 221
Application +4 | -1
Number of values of xx satisfying sin⁡−1(1−x)−2sin⁡−1x=π/2\sin ^{-1}(1-x)-2 \sin ^{-1} x=\pi / 2 is
View answer and explanation
Correct option: B
1
The correct answer is B (1) because after solving the equation, only one value of x satisfies the original equation.
122 Medium
122 of 221
Application +4 | -1
The value of sin⁡(cot⁡−1(cos⁡(tan⁡−1x))\sin \left(\cot ^{-1}\left(\cos \left(\tan ^{-1} x\right)\right)\right. is
View answer and explanation
Correct option: B
x2+1x2+2\sqrt{\frac{x^2+1}{x^2+2}}
Option B (x2+1x2+2\sqrt{\frac{x^2+1}{x^2+2}}) is correct because the expression simplifies using trigonometric identities and right triangle relationships.
123 Medium
123 of 221
Analyzing +4 | -1
The xx satisfying sin⁡−1x+sin⁡−1(1−x)=cos⁡−1x\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x are
View answer and explanation
Correct option: C
0,120, \frac{1}{2}
Option C (0, 1/2) is correct because these are the solutions to the equation obtained after applying trigonometric identities and solving the resulting algebraic equation.
124 Medium
124 of 221
Application +4 | -1
The equation sin⁡−1x=2sin⁡−1a\sin ^{-1} x=2 \sin ^{-1} a holds true for
View answer and explanation
Correct option: C
−12≤a≤12-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}
Option C (−12≤a≤12-\frac{{1}}{{\sqrt{{2}}}} \leq a \leq \frac{{1}}{{\sqrt{{2}}}}) is correct because it defines the range of 'a' for which the equation sin⁡−1x=2sin⁡−1a\sin^{-1} x = 2 \sin^{-1} a holds true, considering the domain and range of inverse trigonometric functions.
125 Medium
125 of 221
Application +4 | -1
tan⁡−1(tan⁡4)+cot⁡−1(cot⁡4)\tan ^{-1}(\tan 4)+\cot ^{-1}(\cot 4) equals
View answer and explanation
Correct option: C
8−2π8-2 \pi
Option C (8−2π8 - 2\pi) is correct because, due to the range restrictions of tan⁡−1\tan^{-1} and cot⁡−1\cot^{-1}, the expression simplifies to 8−2π8 - 2\pi.
126 Medium
126 of 221
Application +4 | -1
If cos⁡−1x+cos⁡−1y+cos⁡−1z=3π\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi then xy+yz+zxx y+y z+z x is equal to :
View answer and explanation
Correct option: D
3
Option D is correct because when cos⁡−1x+cos⁡−1y+cos⁡−1z=3π\cos^{-1}x + \cos^{-1}y + \cos^{-1}z = 3\pi, it implies x=y=z=−1x=y=z=-1, and therefore xy+yz+zx=3xy+yz+zx = 3.
127 Medium
127 of 221
Application +4 | -1
Let x∈(0,1)x \in(0,1). The set of all xx such that sin⁡−1x>cos⁡−1x\sin ^{-1} x>\cos ^{-1} x, is the interval:
View answer and explanation
Correct option: B
(12,1)\left(\frac{1}{\sqrt{2}}, 1\right)
Option B ((12,1)\left(\frac{{1}}{{\sqrt{{2}}}}, 1\right)) is correct because solving the inequality sin⁡−1x>cos⁡−1x\sin ^{-1} x > \cos ^{-1} x leads to x>12x > \frac{{1}}{{\sqrt{{2}}}} and x<1x < 1.
128 Medium
128 of 221
Analyzing +4 | -1
The number of real solutions of the equation tan⁡−1x2−3x+2+cos⁡−14x−x2−3=π\tan ^{-1} \sqrt{x^2-3 x+2}+\cos ^{-1} \sqrt{4 x-x^2-3}=\pi is
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because the given equation has no real solutions as the range of the left-hand side is strictly less than π\pi.
129 Medium
129 of 221
Application +4 | -1
tan⁡[2tan⁡−1(15)−π4]\tan \left[2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right] is equal to
View answer and explanation
Correct option: D
−717-\frac{7}{17}
Option D (−717-\frac{7}{17}) is correct because by applying the double angle formula for tangent and then the tangent subtraction formula, we arrive at the correct value.
130 Medium
130 of 221
Application +4 | -1
Assertion: 2tan⁡−112+tan⁡−117=tan⁡−131172 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}.

Reason: 2tan⁡−1x=tan⁡−1(2x1−x2)2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right) if −1<x<1-1 < x < 1.
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion.
Option A is correct because the assertion is true, the reason is true, and the reason correctly explains the assertion using the trigonometric identity for 2tan⁡−1x2 \tan^{-1}x.
131 Medium
131 of 221
Application +4 | -1
If θ=cot⁡−17+cot⁡−18+cot⁡−118\theta = {\cot ^{ - 1}}7 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18, then the value of cot⁡ θ\cot \,\theta must be
View answer and explanation
Correct option: A
3
Option A (3) is correct because by converting the cotangent inverse to tangent inverse and using the tangent addition formula, we find that cot⁡θ=3\cot \theta = 3.
132 Medium
132 of 221
Application +4 | -1
If a=sin⁡−1(sin⁡(5))a=\sin ^{-1}(\sin (5)) and b=cos⁡−1(cos⁡(5))b=\cos ^{-1}(\cos (5)), then a2+b2a^2+b^2 is equal to
View answer and explanation
Correct option: B
8π2−40π+508 \pi^2-40 \pi+50
Option A (8π2−40π+508 \pi^2-40 \pi+50) is correct because it is the correct simplification after finding a and b using the given conditions.
133 Medium
133 of 221
Analyzing +4 | -1
The equation 2cos⁡−1x+sin⁡−1x=11π62{\cos ^{ - 1}}x + {\sin ^{ - 1}}x = \frac{{11\pi }}{6} has
View answer and explanation
Correct option: A
No solution
Option A (No solution) is correct because the equation leads to a value of cos⁡−1x{\cos ^{ - 1}}x that is outside its defined range of [0, π].
134 Medium
134 of 221
Application +4 | -1
The value of cos⁡[tan⁡−1(tan⁡15π4)]\cos \left[\tan ^{-1}\left(\tan \frac{15 \pi}{4}\right)\right] is
View answer and explanation
Correct option: A
12\frac{1}{\sqrt{2}}
Option A (12\frac{{1}}{{\sqrt{{2}}}}) is correct because it is obtained by simplifying the expression using the properties of tangent and inverse trigonometric functions.
135 Medium
135 of 221
Application +4 | -1
The value of cot⁡−134+sin⁡−1513{\cot ^{ - 1}}\frac{3}{4} + {\sin ^{ - 1}}\frac{5}{{13}} is
View answer and explanation
Correct option: A
sin⁡−16365{\sin ^{ - 1}}\frac{{63}}{{65}}
Option A is correct because by converting both terms to arcsin and using the arcsin addition formula, the expression simplifies to sin⁡−16365{\sin ^{ - 1}}\frac{{63}}{{65}}.

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136 Medium
136 of 221
Application +4 | -1
The value of {tan⁡[π4+12sin⁡−1(ab)]+tan⁡[π4−12sin⁡−1(ab)]}−1\left\{\tan \left[\frac{\pi}{4}+\frac{1}{2} \sin ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \sin ^{-1}\left(\frac{a}{b}\right)\right]\right\}^{-1}, where 0<a<b0 < a < b is
View answer and explanation
Correct option: C
b2−a22b\frac{\sqrt{b^2-a^2}}{2 b}
Option D (b2−a22b\frac{\sqrt{b^2-a^2}}{2 b}) is correct because simplifying the trigonometric expression and using the given condition leads to this result.
137 Medium
137 of 221
Application +4 | -1
If cosec⁡−1x=2cos⁡−17+cos⁡−1(35)\operatorname{cosec}^{-1} x=2 \cos ^{-1} 7+\cos ^{-1}\left(\frac{3}{5}\right) then the value of xx is
View answer and explanation
Correct option: B
125/117125 / 117
Option B (125117\frac{{125}}{{117}}) is correct because it correctly uses the cotangent addition formula and trigonometric identities to find the value of x.
138 Medium
138 of 221
Analyzing +4 | -1
cos⁡−1{12x2+1−x21−x24}=cos⁡−1x2−cos⁡−1x{\cos ^{ - 1}}\left\{ {\frac{1}{2}{x^2} + \sqrt {1 - {x^2}} \sqrt {1 - \frac{{{x^2}}}{4}} } \right\} = {\cos ^{ - 1}}\frac{x}{2} - {\cos ^{ - 1}}x holds for
View answer and explanation
Correct option: A
∣x∣≤1\left| x \right| \le 1
Option A (∣x∣≤1\left| x \right| \le 1) is correct because the domain of the inverse cosine function requires ∣x∣≤1|x| \le 1, and this condition must be satisfied for all terms in the equation.
139 Medium
139 of 221
Application +4 | -1
Considering only the principal values of inverse trigonometric functions, the number of positive real values of xx satisfying tan⁡−1(x)+tan⁡−1(2x)=π4\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4} is:
View answer and explanation
Correct option: B
1
Option B (1) is correct because the function is monotonically increasing, ensuring only one positive real solution for the given equation.
140 Medium
140 of 221
Application +4 | -1
If x1=2tan⁡−1(1+x1−x),x2=sin⁡−1(1−x21+x2)x_1=2 \tan ^{-1}\left(\frac{1+x}{1-x}\right), x_2=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right); where x∈x \in (0,1)(0,1) then x1+x2x_1+x_2 is equal to
View answer and explanation
Correct option: C
π\pi
Option C (π\pi) is correct because simplifying the expression using trigonometric identities and the given condition leads to π\pi.
141 Medium
141 of 221
Analyzing +4 | -1
Sum of series of cot⁡−153+cot⁡−193+cot⁡−1153+cot⁡−1233+……∞\cot ^{-1} \frac{5}{\sqrt{3}}+\cot ^{-1} \frac{9}{\sqrt{3}}+\cot ^{-1} \frac{15}{\sqrt{3}}+\cot ^{-1} \frac{23}{\sqrt{3}}+\ldots \ldots \infty is equal to
View answer and explanation
Correct option: B
π3\frac{\pi}{3}
Option B (π3\frac{\pi}{3}) is correct because the sum of the infinite series of inverse cotangents telescopes to π3\frac{\pi}{3} after converting to inverse tangents and simplifying.
142 Medium
142 of 221
Application +4 | -1
If x>0,y>0x>0, y>0 and x>yx>y, then tan⁡−1(x/y)+tan⁡−1[(x+\tan ^{-1}(x / y)+\tan ^{-1}[(x+ y)/(x−y)]y) /(x-y)] is equal to:
View answer and explanation
Correct option: C
3π/43 \pi / 4
Option C (3π4\frac{{3 \pi}}{{4}}) is correct because the given expression simplifies to π+tan⁡−1(−1)\pi + \tan^{{-1}}(-1) under the given conditions, which equals 3π4\frac{{3\pi}}{{4}}.
143 Medium
143 of 221
Application +4 | -1
The value of the expression sec⁡2(tan⁡−12)+cosec⁡2(cot⁡−13)+cosec⁡(2cot⁡−12+cos⁡−135)\sec ^2\left(\tan ^{-1} 2\right)+\operatorname{cosec}^2\left(\cot ^{-1} 3\right)+\operatorname{cosec}\left(2 \cot ^{-1} 2+\cos ^{-1} \frac{3}{5}\right) is equal to:
View answer and explanation
Correct option: D
15+252415+\frac{25}{24}
Option D (15+252415+\frac{{25}}{{24}}) is correct because simplifying the expression using trigonometric identities and inverse trigonometric functions results in 15+252415+\frac{{25}}{{24}}.
144 Medium
144 of 221
Application +4 | -1
The solution of the inequality (cot⁡−1x)2−5cot⁡−1x+6>0\left(\cot ^{-1} x\right)^2-5 \cot ^{-1} x+6>0 is
View answer and explanation
Correct option: B
(−∞,cot⁡3)∪(cot⁡2,∞)(-\infty, \cot 3) \cup(\cot 2, \infty)
Option B ((−∞,cot⁡3)∪(cot⁡2,∞)(-\infty, \cot 3) \cup (\cot 2, \infty)) is correct because solving the inequality for cot⁡−1x\cot^{-1}x and then finding the corresponding values of xx gives the solution set.
145 Medium
145 of 221
Application +4 | -1
Sum to nn terms of the series
cosec⁡−110+cosec⁡−150+cosec⁡−1170+……+cosec⁡−1(n2+1)(n2+2n+2) is \operatorname{cosec}^{-1} \sqrt{10}+\operatorname{cosec}^{-1} \sqrt{50}+\operatorname{cosec}^{-1} \sqrt{170}+\ldots \ldots \\ +\operatorname{cosec}^{-1} \sqrt{\left(n^2+1\right)\left(n^2+2 n+2\right)} \text { is }
View answer and explanation
Correct option: A
tan⁡−1(n+1)−π4\tan ^{-1}(n+1)-\frac{\pi}{4}
Option A is correct because the sum of the series can be simplified using the arctangent subtraction formula, resulting in tan⁡−1(n+1)−π4\tan^{-1}(n+1) - \frac{\pi}{4}.
146 Medium
146 of 221
Application +4 | -1
If x>0,y>0x > 0, y > 0 and x>yx > y, then tan⁡−1(x/y)+\tan ^{-1}(x / y)+ tan⁡−1[(x+y)/(x−y)]\tan ^{-1}[(x+y) /(x-y)] is equal to
View answer and explanation
Correct option: C
3π/43 \pi / 4
Option C (3π/4) is correct because using the formula for the sum of two arctangents and considering the given conditions, we arrive at the correct answer.
147 Medium
147 of 221
Application +4 | -1
If cot⁡−1(nπ)>π6,n\cot ^{-1}\left(\frac{n}{\pi}\right)>\frac{\pi}{6}, n being a natural number, then maximum value of nn is
View answer and explanation
Correct option: B
5
Option B (5) is correct because solving the inequality cot⁡−1(nπ)>π6\cot^{-1}(\frac{n}{\pi}) > \frac{\pi}{6} leads to n<π3≈5.44n < \pi\sqrt{3} \approx 5.44, and since nn is a natural number, the maximum value is 5.
148 Medium
148 of 221
Application +4 | -1
tan⁡−1(1+33+3)+sec⁡−1(8+436+33)\tan ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+\sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}\right) is equal to:
View answer and explanation
Correct option: A
π3\frac{\pi}{3}
The correct answer is A (π3\frac{{\pi}}{{3}}) because it is the sum of the simplified inverse trigonometric functions.
149 Medium
149 of 221
Application +4 | -1
Solution of equation cos⁡−1x3+cos⁡−1x=π2\cos ^{-1} x \sqrt{3}+\cos ^{-1} x=\frac{\pi}{2} is
View answer and explanation
Correct option: A
x=12x=\frac{1}{2}
Option A (x=12x=\frac{1}{2}) is correct because it satisfies the given equation after simplification and considering the domain of inverse cosine function.
150 Medium
150 of 221
Application +4 | -1
If cot⁡−1α+cot⁡−1β=cot⁡−1x\cot ^{-1} \alpha+\cot ^{-1} \beta=\cot ^{-1} x, then xx is equal to
View answer and explanation
Correct option: D
αβ−1α+β\frac{\alpha \beta-1}{\alpha+\beta}
Option D (αβ−1α+β\frac{{\alpha \beta-1}}{{\alpha+\beta}}) is correct because it is derived from the cotangent addition formula applied to the given equation.
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