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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Easy

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Practice format and chapter quiz

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Showing 61-90 of 365 questions
61 Easy
61 of 365
Application +4 | -1
If tan⁡−1(ax)+tan⁡−1(bx)=π2,{\tan ^{ - 1}}\left( {\frac{a}{x}} \right) + {\tan ^{ - 1}}\left( {\frac{b}{x}} \right) = \frac{\pi }{2}, then xx is equal to
View answer and explanation
Correct option: A
ab\sqrt {ab}
Option A (ab\sqrt {ab}) is correct because solving the given equation using trigonometric identities leads to x=abx = \sqrt{ab}.
62 Easy
62 of 365
Application +4 | -1
The value of cos⁡−1(cos⁡2π3)−sin⁡−1(sin⁡2π3)\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)-\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right) is equal to
View answer and explanation
Correct option: D
π3\frac{\pi}{3}
Option D (π3\frac{\pi}{3}) is correct because it is the simplified result after evaluating the inverse trigonometric functions and subtracting.
63 Easy
63 of 365
Application +4 | -1
Two angles of a triangle are cot⁡−12\cot ^{-1} 2 and cot⁡−13\cot ^{-1} 3, then the third angle are -
View answer and explanation
Correct option: B
3π4\frac{3 \pi}{4}
Option B (3π4\frac{3 \pi}{4}) is correct because the sum of the two given angles is π4\frac{\pi}{4}, and the sum of all angles in a triangle is π\pi, so the third angle is π−π4=3π4\pi - \frac{\pi}{4} = \frac{3 \pi}{4}.
64 Easy
64 of 365
Analyzing +4 | -1
cot⁡−1(2.12)+cot⁡−1(2.22)+cot⁡−1(2.32)+...{\cot ^{ - 1}}\left( {{{2.1}^2}} \right) + {\cot ^{ - 1}}\left( {{{2.2}^2}} \right) + {\cot ^{ - 1}}\left( {{{2.3}^2}} \right) + ... upto ∞\infty is equal to
View answer and explanation
Correct option: A
π4\frac{\pi }{4}
Option A (π4\frac{\pi }{4}) is correct because the series can be expressed as a telescoping sum of arctangents, which simplifies to π4\frac{\pi }{4}.
65 Easy
65 of 365
Analyzing +4 | -1
If a,  b,  ca,\;b,\;c be positive real number and the value of

θ=tan⁡−1a(a+b+c)bc+tan⁡−1b(a+b+c)ca+tan⁡−1c(c+b+c)ab\mathrm{\theta } = {\tan ^{ - 1}}\sqrt {\frac{{a\left( {a + b + c} \right)}}{{bc}}} + {\tan ^{ - 1}}\sqrt {\frac{{b\left( {a + b + c} \right)}}{{ca}}} + {\tan ^{ - 1}}\sqrt {\frac{{c\left( {c + b + c} \right)}}{{ab}}}

Then tan θ\mathrm{\theta } is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because after substituting and simplifying, the expression inside the tangent function becomes zero, resulting in tan θ = 0.
66 Easy
66 of 365
Application +4 | -1
If −π2<sin⁡−1x<π2- \frac{\pi }{2} < {\sin ^{ - 1}}x < \frac{\pi }{2}, then tan⁡(sin⁡−1x)\tan \left( {{{\sin }^{ - 1}}x} \right) is equal to
View answer and explanation
Correct option: C
x1−x2\frac{x}{{\sqrt {1 - {x^2}} }}
Option C (x1−x2\frac{x}{{\sqrt {1 - {x^2}} }}) is correct because it represents the tangent of the angle whose sine is x, simplified using trigonometric identities.
67 Easy
67 of 365
Application +4 | -1
If α,  β\alpha ,\;\beta are the roots of the equation 6x2−5x+1=06{x^2} - 5x + 1 = 0, then the value of tan⁡−1α+tan⁡−1β{\tan ^{ - 1}}\alpha + {\tan ^{ - 1}}\beta is
View answer and explanation
Correct option: B
π/4\pi /4
Option B (π/4\pi /4) is correct because by finding the sum and product of the roots and using the formula for tan⁡−1α+tan⁡−1β{\tan ^{ - 1}}\alpha + {\tan ^{ - 1}}\beta, we can simplify the expression to tan⁡−1(1){\tan ^{ - 1}}(1), which equals π4\frac{\pi}{4}.
68 Easy
68 of 365
Application +4 | -1
If x≥1x \ge 1, then 2tan⁡−1x+sin⁡−1(2x1+x2)2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) is equal to
View answer and explanation
Correct option: D
π\pi
Option D is correct because for x≥1x \ge 1, sin⁡−1(2x1+x2)=π−2tan⁡−1x{\sin ^{ - 1}}(\frac{{2x}}{{1 + {x^2}}}) = \pi - 2{\tan ^{ - 1}}x, and thus the expression simplifies to π\pi.
69 Easy
69 of 365
Remembering +4 | -1
Prinicpal value of tan⁡−1(3)\tan ^{-1}(\sqrt{3}) is equal to
View answer and explanation
Correct option: B
π3\frac{\pi}{3}
Option B (π3\frac{\pi}{3}) is correct because the principal value of tan⁡−1(3)\tan^{-1}(\sqrt{3}) lies in the range (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) and tan⁡(π3)=3\tan(\frac{\pi}{3}) = \sqrt{3}.
70 Easy
70 of 365
Analyzing +4 | -1
The number of real solution of tan⁡−1x(x+1)+sin⁡−1  x2+x+1=π2{\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\;\sqrt {{x^2} + x + 1} = \frac{\pi }{2} is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the equation is satisfied for two values of x, namely x = 0 and x = -1.
71 Easy
71 of 365
Application +4 | -1
The value of cos⁡[tan⁡−1{sin⁡(cot⁡−1x)}]\cos \left[ {{{\tan }^{ - 1}}\left\{ {\sin \left( {{{\cot }^{ - 1}}x} \right)} \right\}} \right] is
View answer and explanation
Correct option: D
x2+1x2+2\sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}}
Option D is correct because simplifying the expression using trigonometric identities leads to x2+1x2+2\sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}}.
72 Easy
72 of 365
Analyzing +4 | -1
Which one of the following is correct?
View answer and explanation
Correct option: A
tan⁡1>tan⁡−11\tan 1 > {\tan ^{ - 1}}1
Option A (tan⁡1>tan⁡−11\tan 1 > {\tan ^{ - 1}}1) is correct because tan⁡1\tan 1 (where 1 is in radians) is greater than 1, while tan⁡−11=π4\tan^{-1} 1 = \frac{\pi}{4} which is less than 1.
73 Easy
73 of 365
Application +4 | -1
If sin⁡−1(x5)+cosec−1(54)=π2{\sin ^{ - 1}}\left( {\frac{x}{5}} \right) + \mathrm{cose}{\mathrm{c}^{ - 1}}\left( {\frac{5}{4}} \right) = \frac{\pi }{2}, then value of xx is
View answer and explanation
Correct option: B
3
Option B (3) is correct because by using the property that sin⁡−1x+csc⁡−1y=π2{\sin ^{ - 1}}x + {\csc ^{ - 1}}y = \frac{\pi }{2} implies x=1yx = \frac{1}{y}, we can solve for x.
74 Easy
74 of 365
Application +4 | -1
Let a1=1,a2,a3,a4,…a_1=1, a_2, a_3, a_4, \ldots be consecutive natural numbers. Then

tan⁡−1(11+a1a2)+tan⁡−1(11+a2a3)+…+tan⁡−1(11+a2021a2022) is equal to \tan ^{-1}\left(\frac{1}{1+a_1 a_2}\right) +\tan ^{-1}\left(\frac{1}{1+a_2 a_3}\right)+\ldots +\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right) \text { is equal to }
View answer and explanation
Correct option: A
tan⁡−1(2022)−π4\tan ^{-1}(2022)-\frac{\pi}{4}
Option A (tan⁡−1(2022)−π4\tan ^{-1}(2022)-\frac{{\pi}}{{4}}) is correct because the given sum telescopes to tan⁡−1(a2022)−tan⁡−1(a1)=tan⁡−1(2022)−tan⁡−1(1)=tan⁡−1(2022)−π4\tan^{-1}(a_{2022}) - \tan^{-1}(a_1) = \tan^{-1}(2022) - \tan^{-1}(1) = \tan^{-1}(2022) - \frac{{\pi}}{{4}}.
75 Easy
75 of 365
Application +4 | -1
If −1≤x≤−12- 1 \le x \le - \frac{1}{{\sqrt 2 }}, then sin⁡−1(2x1−x2){\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) equals
View answer and explanation
Correct option: C
−π−2sin⁡−1x- \pi - 2{\sin ^{ - 1}}x
Option C (−π−2sin⁡−1x- \pi - 2{\sin ^{ - 1}}x) is correct because, within the given range for x, the expression sin⁡−1(2x1−x2){\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ) simplifies to this value after considering the range of the inverse sine function.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
76 Easy
76 of 365
Application +4 | -1
If cos⁡−1(23x)+cos⁡−1(34x)=π2(x>34)\cos ^{-1}\left(\frac{2}{3 x}\right)+\cos ^{-1}\left(\frac{3}{4 x}\right)=\frac{\pi}{2}\left(x>\frac{3}{4}\right), then xx is equal to:
View answer and explanation
Correct option: D
14512\frac{\sqrt{145}}{12}
Option D (14512\frac{\sqrt{{145}}}{{12}}) is correct because solving the given equation using trigonometric identities leads to x=14512x = \frac{\sqrt{{145}}}{{12}}.
77 Easy
77 of 365
Application +4 | -1
The relation tan⁡−1(1+x1−x)=π4+tan⁡−1x{\tan ^{ - 1}}\left( {\frac{{1 + x}}{{1 - x}}} \right) = \frac{\pi }{4} + {\tan ^{ - 1}}x holds true for all
View answer and explanation
Correct option: B
x∈(−∞,  1)x \in \left( { - \infty ,\;1} \right)
Option B (x∈(−∞,  1)x \in \left( { - \infty ,\;1} \right)) is correct because the identity holds true only when x<1x < 1 due to the domain restrictions of the inverse tangent function.
78 Easy
78 of 365
Analyzing +4 | -1
If x>13x > \frac{1}{{\sqrt 3 }}, then x>13x > \frac{1}{{\sqrt 3 }} equals
View answer and explanation
Correct option: B
−π+3tan⁡−1x- \pi + 3{\tan ^{ - 1}}x
Option B (−π+3tan⁡−1x- \pi + 3{\tan ^{ - 1}}x) is correct because it represents the correct simplification of the given expression when x>13x > \frac{1}{{\sqrt 3 }}.
79 Easy
79 of 365
Application +4 | -1
If cos⁡−1x>sin⁡−1x,  then{\cos ^{ - 1}}x > {\sin ^{ - 1}}x,\;\mathrm{then}
View answer and explanation
Correct option: D
−1≤x<12- 1 \le x < \frac{1}{{\sqrt 2 }}
Option D (−1≤x<12- 1 \le x < \frac{1}{{\sqrt 2 }}) is correct because it represents the range of x where the inverse cosine is greater than the inverse sine.
80 Easy
80 of 365
Application +4 | -1
If 3+i=(a+ib)(c+id)\sqrt 3 + i = \left( {a + ib} \right)\left( {c + id} \right), then tan⁡−1(ba)+tan⁡−1(dc){\tan ^{ - 1}}\left( {\frac{b}{a}} \right) + {\tan ^{ - 1}}\left( {\frac{d}{c}} \right) has the value
View answer and explanation
Correct option: B
nπ+π6,  n∈In\pi + \frac{\pi }{6},\;n \in I
Option B (nπ+π6,  n∈In\pi + \frac{\pi }{6},\;n \in I) is correct because it is derived by using the tangent addition formula and simplifying the given expression.
81 Easy
81 of 365
Application +4 | -1
Let x∗y=x2+y3{x^*}y = {x^2} + {y^3} and (x∗1)∗1=x∗(1∗1).{({x^*}1)^*}1 = {x^*}({1^*}1). Then a value of 2sin⁡−1(x4+x2−2x4+x2+2)2{\sin ^{ - 1}}\left( {\frac{{{x^4} + {x^2} - 2}}{{{x^4} + {x^2} + 2}}} \right) is
View answer and explanation
Correct option: B
π3\frac{\pi }{3}
Option B (π3\frac{\pi}{3}) is correct because by substituting the given operation into the equation and solving for x, we can find the value of the expression.
82 Easy
82 of 365
Application +4 | -1
If cosec⁡−1x=2cot⁡−17+cos⁡−1\operatorname{cosec}^{-1} x=2 \cot ^{-1} 7+\cos ^{-1} (3/5) then the value of xx is
View answer and explanation
Correct option: B
125/117125 / 117
Option B (125/117) is correct because, by applying the cotangent addition formula and simplifying, we find that cosec-1x = cosec-1(125/117), thus x = 125/117.
83 Easy
83 of 365
Application +4 | -1
sin⁡[π3−sin⁡−1(−12)]\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right] is equal to
View answer and explanation
Correct option: D
1
Option D (1) is correct because it is the value of the sine function after simplifying the expression using the inverse sine function.
84 Easy
84 of 365
Application +4 | -1
If sin⁡−12a1+a2−cos⁡−11−b21+b2=tan⁡−12x1−x2,{\sin ^{ - 1}}\frac{{2a}}{{1 + {a^2}}} - {\cos ^{ - 1}}\frac{{1 - {b^2}}}{{1 + {b^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}, then value of xx is
View answer and explanation
Correct option: D
a−b1+ab\frac{{a - b}}{{1 + ab}}
Option A (0) is correct because when sin⁡−1x+sin⁡−1y+sin⁡−1z=3π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{3\pi }}{2}, it implies x=y=z=1, and the expression evaluates to 0.
85 Easy
85 of 365
Application +4 | -1
cot⁡(π4−2cot⁡−13)=\cot \left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)=
View answer and explanation
Correct option: A
7
Option A (7) is correct because by applying trigonometric identities and simplification, the expression evaluates to 7.
86 Easy
86 of 365
Application +4 | -1
tan⁡−13−sec⁡−1(−2)\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2) is equal to
View answer and explanation
Correct option: B
−π3-\frac{\pi}{3}
Option B (−π3-\frac{{\pi}}{{3}}) is correct because it is the result of subtracting the principal values of the inverse secant from the inverse tangent.
87 Easy
87 of 365
Application +4 | -1
The value of xx for which cos⁡−1(cos⁡4)>3x2−4x  {\cos ^{ - 1}}(\cos 4) > 3{x^2} - 4x\; is
View answer and explanation
Correct option: D
(2−6π−83,2+6π−83)\left( {\frac{{2 - \sqrt {6\pi - 8} }}{3},\frac{{2 + \sqrt {6\pi - 8} }}{3}} \right)
Option D ((2−6π−83,2+6π−83)\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){\cos ^{ - 1}}(\cos 4).
88 Easy
88 of 365
Application +4 | -1
cot⁡−1(−12)+cot⁡−1(−13)\cot ^{-1}\left(-\frac{1}{2}\right)+\cot ^{-1}\left(-\frac{1}{3}\right) is equal to
View answer and explanation
Correct option: B
5π4\frac{5 \pi}{4}
Option B (5π4\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.
89 Easy
89 of 365
Application +4 | -1
A root of the equation 17x2+17xtan⁡[2tan⁡−1(1/5)−π/4]−10=017 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.
90 Easy
90 of 365
Analyzing +4 | -1
Let f(x)=(sin⁡(tan⁡−1x)+sin⁡(cot⁡−1x))2−1, ∣x∣>1.    If  dydx=12ddx(sin⁡−1(f(x)))    and   y(3)=π6f\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(−3)y\left( { - \sqrt 3 } \right) is equal to
View answer and explanation
Correct option: B
−π6- \frac{\pi }{6}
Option B (−π6\frac{{ - \pi }}{6}) is correct because after simplifying f(x)f(x) and using the given initial condition, we can find y(−3)y(-\sqrt{3}).
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