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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Easy

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91 Easy
91 of 365
Application +4 | -1
If θ=sin⁡−1x+cos⁡−1x−tan⁡−1x≥0,\mathrm{\theta } = {\sin ^{ - 1}}x + {\cos ^{ - 1}}x - {\tan ^{ - 1}}x \ge 0, then the smallest interval in which θ\mathrm{\theta } lies, is given by
View answer and explanation
Correct option: D
π4≤θ≤π2\frac{\pi }{4} \le \mathrm{\theta } \le \frac{\pi }{2}
Option D (π4≤θ≤π2\frac{\pi }{4} \le \mathrm{\theta } \le \frac{\pi }{2}) is correct because, given θ=sin⁡−1x+cos⁡−1x−tan⁡−1x≥0\mathrm{\theta } = {\sin ^{ - 1}}x + {\cos ^{ - 1}}x - {\tan ^{ - 1}}x \ge 0, the smallest interval in which θ\mathrm{\theta } lies is π4≤θ≤π2\frac{\pi }{4} \le \mathrm{\theta } \le \frac{\pi }{2}.
92 Easy
92 of 365
Application +4 | -1
The value of tan⁡−1[1+x2+1−x21+x2−1−x2],∣x∣<12,x≠0\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
π4+12cos⁡−1x2\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} \mathrm{x}^{2}
The correct answer is A (π4+12cos⁡−1x2\frac{\pi}{4}+\frac{1}{2} \cos^{-1} \mathrm{x}^{2}) because it involves using trigonometric substitution and simplification.
93 Easy
93 of 365
Application +4 | -1
If sin⁡−1(x−x22+x34+….)+cos⁡−1(x2−x42+x64+….)=π2\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∣<20 < |x| < \sqrt{2} then xx 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.
94 Easy
94 of 365
Application +4 | -1
The principal value of sin⁡−1[sin⁡(2π3)]\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 π3\frac{{\pi}}{{3}}.
95 Easy
95 of 365
Application +4 | -1
If sin⁡−1x+sin⁡−1(1−x)=cos⁡−1x{\sin ^{ - 1}}x + {\sin ^{ - 1}}\left( {1 - x} \right) = {\cos ^{ - 1}}x, then xx belongs to
View answer and explanation
Correct option: C
{0,12}\left\{ {0,\frac{1}{2}} \right\}
Option C ({0,12}\left\{ {0,\frac{1}{2}} \right\}) is correct because these are the solutions obtained after simplifying the given equation and checking for validity.
96 Easy
96 of 365
Analyzing +4 | -1
If α=sin⁡−145+sin⁡−113and  β=cos⁡−145+cos⁡−113,  \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
α<β\mathrm{\alpha } < \beta
Option A (α<β\alpha < \beta) is correct because by calculating the values of α\alpha and β\beta, we can show that α\alpha is less than β\beta.
97 Easy
97 of 365
Application +4 | -1
sin⁡−1(sin⁡2π3)+cos⁡−1(cos⁡7π6)+tan⁡−1(tan⁡3π4){\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
11π12\frac{{11\pi }}{{12}}
Option A (11π/12) is correct because it involves evaluating inverse trigonometric functions and simplifying the result.
98 Easy
98 of 365
Application +4 | -1
The value of sin⁡−1(32)+sin⁡−1(12)\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∘105^{\circ}
Option B (105∘105^{\circ}) is correct because sin⁡−1(32)+sin⁡−1(12)\sin^{-1}(\frac{\sqrt{3}}{2}) + \sin^{-1}(\frac{1}{\sqrt{2}}) equals π3+π4\frac{\pi}{3} + \frac{\pi}{4}, which simplifies to 7π12\frac{7\pi}{12} or 105∘105^{\circ}.
99 Easy
99 of 365
Application +4 | -1
If −1≤x≤0- 1 \le x \le 0, then cos⁡−1(2x2−1){\cos ^{ - 1}}\left( {2{x^2} - 1} \right) equals
View answer and explanation
Correct option: C
2  π−2cos⁡−1x2\;\pi - 2{\cos ^{ - 1}}x
The correct answer is C (2  π−2cos⁡−1x2\;\pi - 2{\cos ^{ - 1}}x) because considering the given range of x, cos⁡−1(2x2−1){\cos ^{ - 1}}(2x^2 - 1) simplifies to 2π−2cos⁡−1x2\pi - 2{\cos ^{ - 1}}x.
100 Easy
100 of 365
Understanding +4 | -1
The value of sin⁡−1(32)−sin⁡−1(12){\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) - {\sin ^{ - 1}}\left( {\frac{1}{2}} \right) is
View answer and explanation
Correct option: D
30∘30^\circ
Option D (30∘30^\circ) is correct because sin⁡−1(32)=60∘{\sin ^{ - 1}}(\frac{{\sqrt 3 }}{2}) = 60^\circ and sin⁡−1(12)=30∘{\sin ^{ - 1}}(\frac{1}{2}) = 30^\circ, so the difference is 60∘−30∘=30∘60^\circ - 30^\circ = 30^\circ.
101 Easy
101 of 365
Application +4 | -1
If sin⁡−1x+sin⁡−1y+sin⁡−1z=−3π2\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=-\frac{3 \pi}{2} then the value of xy+yz+zxx y+y z+z x is
View answer and explanation
Correct option: A
3
The correct answer is A (3) 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 therefore xy + yz + zx = 3.
102 Easy
102 of 365
Analyzing +4 | -1
If cos⁡−1x=α,  (0<x<1){\cos ^{ - 1}}x = \alpha ,\;(0 < x < 1) and

sin⁡−1(2x1−x2)+sec⁡−1(12x2−1)=2π3{\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(2x){\tan ^{ - 1}}\left( {2x} \right) equals
View answer and explanation
Correct option: C
π/3\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.
103 Easy
103 of 365
Application +4 | -1
The value of sin⁡(4tan⁡−113)−cos⁡(2tan⁡−117)\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.
104 Easy
104 of 365
Application +4 | -1
If cos⁡−1p+cos⁡−11−p+cos⁡−11−q=3π4,{\cos ^{ - 1}}\sqrt p + {\cos ^{ - 1}}\sqrt {1 - p} + {\cos ^{ - 1}}\sqrt {1 - q} = \frac{{3\pi }}{4}, then the value of qq is
View answer and explanation
Correct option: D
12\frac{1}{2}
Option D (12\frac{1}{2}) is correct because by using the given equation and trigonometric identities, we can derive that q=12q = \frac{1}{2}.
105 Easy
105 of 365
Application +4 | -1
A solution of the equation tan⁡−1(1+x)+tan⁡−1(1−x)=π2,{\tan ^{ - 1}}\left( {1 + x} \right) + {\tan ^{ - 1}}\left( {1 - x} \right) = \frac{\pi }{2}, is
View answer and explanation
Correct option: C
x=0x = 0
Option C is correct because solving the equation tan⁡−1(1+x)+tan⁡−1(1−x)=π2{\tan ^{ - 1}}\left( {1 + x} \right) + {\tan ^{ - 1}}\left( {1 - x} \right) = \frac{\pi }{2} using trigonometric identities leads to x=0x = 0.

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106 Easy
106 of 365
Application +4 | -1
If tan⁡(x+y)=33\tan (x+y)=33 and x=tan⁡−13x=\tan ^{-1} 3, the value of yy is
View answer and explanation
Correct option: A
0.3
Option B (tan⁡−1(0.3)\tan^{-1}(0.3)) is correct because by using the tangent addition formula, we find that tan⁡y=0.3\tan y = 0.3, so y=tan⁡−1(0.3)y = \tan^{-1}(0.3).
107 Easy
107 of 365
Remembering +4 | -1
Principal value of sec⁡−1(2)\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 the principal value of sec⁡−1(2)\sec^{-1}(2) lies in the range [0,π]−π2[0, \pi] - {\frac{\pi}{2}} and sec⁡(π3)=2\sec(\frac{\pi}{3}) = 2.
108 Easy
108 of 365
Analyzing +4 | -1
If x∈(−π2,π2),x \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right), then the value of

tan⁡−1(tan⁡x4)+tan⁡−1(3sin⁡2x5+3cos⁡2x){\tan ^{ - 1}}\left( {\frac{{\tan x}}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{{3\sin 2x}}{{5 + 3\cos 2x}}} \right) is
View answer and explanation
Correct option: D
xx
Option A (No solution) is correct because the equation leads to cos⁡−1x=4π3{\cos ^{ - 1}}x = \frac{{4\pi }}{3}, which is not possible since the range of cos⁡−1x{\cos ^{ - 1}}x is [0,π][0, \pi].
109 Easy
109 of 365
Application +4 | -1
The number of solutions of the equation sin⁡−1x+sin⁡−12x=π3{\sin ^{ - 1}}x + {\sin ^{ - 1}}2x = \frac{\pi }{3}, is
View answer and explanation
Correct option: B
1
Option B (1) is correct because, after solving the equation, only one valid solution remains within the domain of the inverse sine functions.
110 Easy
110 of 365
Application +4 | -1
If 12≤x≤1\frac{1}{2} \le x \le 1, then cos⁡−1(4x3−3x)  {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)\mathrm{\;} equals
View answer and explanation
Correct option: A
3cos⁡−1x3{\cos ^{ - 1}}x
Option A (3cos⁡−1x3{\cos ^{ - 1}}x) is correct because by substituting x=cos⁡θx = \cos \theta, the expression simplifies to cos⁡−1(cos⁡3θ){\cos ^{ - 1}}(\cos 3\theta), which equals 3cos⁡−1x3{\cos ^{ - 1}}x within the given interval.
111 Easy
111 of 365
Application +4 | -1
sin⁡−1(1−x)−2sin⁡−1x=π/2\sin ^{-1}(1-x)-2 \sin ^{-1} x=\pi / 2, then xx is equal to
View answer and explanation
Correct option: C
00
Option C (0) is correct because after solving the equation, we get two possible solutions, 0 and 1/2. However, only 0 satisfies the original equation.
112 Easy
112 of 365
Application +4 | -1
The value of tan⁡−1(cos⁡(15π4)−1sin⁡(π4)){\tan ^{ - 1}}\left( {\frac{{\cos \left( {\frac{{15\pi }}{4}} \right) - 1}}{{\sin \left( {\frac{\pi }{4}} \right)}}} \right) is equal to:
View answer and explanation
Correct option: B
−π8- \frac{\pi }{8}
Option B (−π8\frac{-\pi}{8}) is correct because it is the simplified value of the given inverse trigonometric expression after evaluating the trigonometric functions.
113 Easy
113 of 365
Application +4 | -1
The value of cos⁡−1(cos⁡12)−sin⁡−1(sin⁡14){\cos ^{ - 1}}\left( {\cos 12} \right) - {\sin ^{ - 1}}\left( {\sin 14} \right) is
View answer and explanation
Correct option: B
8π−268\pi - 26
Option B (8π−268\pi - 26) is correct because it is the simplified value of the given expression after considering the range of inverse trigonometric functions and periodicity.
114 Easy
114 of 365
Application +4 | -1
If a1,a2,a3,…an{a_1},{a_2},{a_3}, \ldots {a_n} are in AP with common difference 5 and if aiaj≠−1{a_i}{a_j} \ne - 1 for i,  j=1,  2,  …,ni,\;j = 1,\;2,\; \ldots ,n then

tan⁡−1(51+a1a2)+tan⁡−1(51+a2a3)+...+tan⁡−1(51+an−1an){\tan ^{ - 1}}\left( {\frac{5}{{1 + {a_1}{a_2}}}} \right) + {\tan ^{ - 1}}\left( {\frac{5}{{1 + {a_2}{a_3}}}} \right) + ... + {\tan ^{ - 1}}\left( {\frac{5}{{1 + {a_{n - 1}}{a_n}}}} \right) is equal to
View answer and explanation
Correct option: C
tan⁡−1(5n−51+ana1){\tan ^{ - 1}}\left( {\frac{{5n - 5}}{{1 + {a_n}{a_1}}}} \right)
Option C (tan⁡−1(5n−51+ana1){\tan ^{ - 1}}\left( {\frac{{5n - 5}}{{1 + {a_n}{a_1}}}} \right)) is correct because by using the properties of arithmetic progressions and inverse trigonometric functions, the sum simplifies to the given expression.
115 Easy
115 of 365
Application +4 | -1
If tanh⁡−1(x+iy)=12tanh⁡−1(2x1+x2+y2)+i2tan⁡−1(2y1−x2−y2),   x,   y∈R{\tanh ^{ - 1}}\left( {x + iy} \right) = \frac{1}{2}{\tanh ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2} + {y^2}}}} \right) + \frac{i}{2}{\tan ^{ - 1}}\left( {\frac{{2y}}{{1 - {x^2} - {y^2}}}} \right),\;\,x,\;\,y \in R, then tanh⁡−1(iy){\tanh ^{ - 1}}\left( {iy} \right) is
View answer and explanation
Correct option: C
itan⁡−1yi{\tan ^{ - 1}}y
Option C (itan⁡−1yi{\tan ^{ - 1}}y) is correct because it is derived by substituting x=0x = 0 into the given equation and simplifying using trigonometric identities.
116 Easy
116 of 365
Application +4 | -1
Which one of following is true?
View answer and explanation
Correct option: A
sin⁡(cos⁡−1x)=cos⁡(sin⁡−1x)\sin \left( {{{\cos }^{ - 1}}x} \right) = \cos \left( {{{\sin }^{ - 1}}x} \right)
Option A is correct because sin⁡(cos⁡−1x)\sin({{\cos }^{ - 1}}x) and cos⁡(sin⁡−1x)\cos({{\sin }^{ - 1}}x) both simplify to the same expression, 1−x2\sqrt{1-x^2}.
117 Easy
117 of 365
Application +4 | -1
The solution of tan⁡−12θ+tan⁡−13θ=π4{\tan ^{ - 1}}2\mathrm{\theta } + {\tan ^{ - 1}}3\mathrm{\theta } = \frac{\pi }{4} is
View answer and explanation
Correct option: C
16\frac{1}{6}
Option C (16\frac{1}{6}) is correct because solving the equation tan⁡−12θ+tan⁡−13θ=π4{\tan ^{ - 1}}2\theta + {\tan ^{ - 1}}3\theta = \frac{\pi }{4} using the tangent addition formula yields θ=16\theta = \frac{1}{6} as the valid solution.
118 Easy
118 of 365
Application +4 | -1
If sin⁡−1x+sin⁡−12x=π/3\sin ^{-1} x+\sin ^{-1} 2 x=\pi / 3, then x=x=
View answer and explanation
Correct option: B
(1/2)(3/7)(1 / 2) \sqrt{(3 / 7)}
The correct answer is B ((1/2)(3/7)(1 / 2) \sqrt{(3 / 7)}) because solving the equation using trigonometric identities and checking for valid solutions yields this result.
119 Easy
119 of 365
Application +4 | -1
Considering only the principal values of the inverse trigonometric functions, the domain of the function f(x)=cos⁡−1(x2−4x+2x2+3)f(x)=\cos ^{-1}\left(\frac{x^2-4 x+2}{x^2+3}\right) is
View answer and explanation
Correct option: B
[−14,∞)\left[-\frac{1}{4}, \infty\right)
Option B ([−14,∞)\left[-\frac{1}{4}, \infty\right)) is correct because it represents the domain where the argument of the inverse cosine function lies between -1 and 1.
120 Easy
120 of 365
Application +4 | -1
The value of (1/2)cos⁡−1(3/5)(1 / 2) \cos ^{-1}(3 / 5) is
View answer and explanation
Correct option: D
tan⁡−1(1/2)\tan ^{-1}(1 / 2)
Option D, tan⁡−1(1/2)\tan^{-1}(1/2), is correct because by using the double angle formula for cosine, we can derive the tangent of half the angle.
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