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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Mixed difficulty

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Showing 181-210 of 605 questions

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181 Medium
181 of 605
Application +4 | -1
If sin⁡−1x−cos⁡−1x=π6,{\sin ^{ - 1}}x - {\cos ^{ - 1}}x = \frac{\pi }{6}, then xx is
View answer and explanation
Correct option: B
32\frac{{\sqrt 3 }}{2}
Option B (32\frac{{\sqrt 3 }}{2}) is correct because solving the equation using trigonometric identities leads to this value for x.
182 Medium
182 of 605
Application +4 | -1
If tan⁡−1(a/x)+tan⁡−1(b/x)=π/2\tan ^{-1}(a / x)+\tan ^{-1}(b / x)=\pi / 2, then xx is equal to
View answer and explanation
Correct option: D
ab\sqrt{a b}
Option D, ab\sqrt{ab}, is correct because by applying the tangent addition formula and using the given condition, we can derive that x=abx = \sqrt{ab}.
183 Easy
183 of 605
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}.
184 Medium
184 of 605
Application +4 | -1
If cos⁡(2sin⁡−1x)=19\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}, then xx is equal to
View answer and explanation
Correct option: C
23,−23\frac{2}{3},-\frac{2}{3}
Option C (23,−23\frac{2}{3}, -\frac{2}{3}) is correct because solving the trigonometric equation cos⁡(2sin⁡−1x)=19\cos(2\sin^{-1}x) = \frac{1}{9} yields both positive and negative solutions for x.
185 Easy
185 of 605
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.
186 Easy
186 of 605
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).
187 Medium
187 of 605
Analyzing +4 | -1
Sum of infinite terms of the series

cot⁡−1(12+34)+cot⁡−1(22+34)+cot⁡−1(32+34)+...{\cot ^{ - 1}}\left( {{1^2} + \frac{3}{4}} \right) + {\cot ^{ - 1}}\left( {{2^2} + \frac{3}{4}} \right) + {\cot ^{ - 1}}\left( {{3^2} + \frac{3}{4}} \right) + ... is
View answer and explanation
Correct option: B
tan⁡−1(2){\tan ^{ - 1}}\left( 2 \right)
Option B (tan⁡−1(2){\tan ^{ - 1}}( 2 )) is correct because the series telescopes to tan⁡−1(n+12)−tan⁡−1(n−12){\tan ^{ - 1}}(n + \frac{1}{2}) - {\tan ^{ - 1}}(n - \frac{1}{2}) and sums to tan⁡−1(2){\tan ^{ - 1}}(2).
188 Medium
188 of 605
Application +4 | -1
If x∈(3π2,2π)x \in\left(\frac{3 \pi}{2}, 2 \pi\right) then value of the expression sin⁡−1(cos⁡(cos⁡−1(cos⁡x)+sin⁡−1(sin⁡x)))\sin ^{-1}\left(\cos \left(\cos ^{-1}(\cos x)+\sin ^{-1}(\sin x)\right)\right) equals
View answer and explanation
Correct option: B
π2\frac{\pi}{2}
Option B (π2\frac{\pi}{2}) is correct because the expression simplifies to sin⁡−1(1)\sin^{-1}(1), which equals π2\frac{\pi}{2}.
189 Medium
189 of 605
Application +4 | -1
The value of cos⁡−1x+cos⁡−1(x2+123−3x2)(1/2≤x≤1)\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right) (1 / 2 \leq x \leq 1) is equal to
View answer and explanation
Correct option: B
π/3\pi / 3
Option B (π/3\pi / 3) is correct because the expression simplifies to cos⁡−1x+(π/3−cos⁡−1x)\cos^{-1}x + (\pi/3 - \cos^{-1}x), which equals π/3\pi/3.
190 Easy
190 of 605
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.
191 Easy
191 of 605
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].
192 Medium
192 of 605
Application +4 | -1
If cos⁡−1[sin⁡(sin⁡−1(sin⁡x))]=0\cos ^{-1}\left[\sin \left(\sin ^{-1}(\sin x)\right)\right]=0, then the value of tan⁡x2\tan \frac{x}{2} is (x∈[0,2π])(x \in[0,2 \pi])
View answer and explanation
Correct option: A
1
Option A (1) is correct because the equation simplifies to sin x = 1, which implies x = π/2, and thus tan(x/2) = tan(π/4) = 1.
193 Easy
193 of 605
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.
194 Medium
194 of 605
Application +4 | -1
If tan⁡−1x=2cot⁡−17+cos⁡−1(35)\tan ^{-1} x=2 \cot ^{-1} 7+\cos ^{-1}\left(\frac{3}{5}\right), then x=x=
View answer and explanation
Correct option: A
11744\frac{117}{44}
Option A (11744\frac{{117}}{{44}}) is the correct answer because it is obtained by simplifying the given trigonometric equation using the tangent addition formula.
195 Easy
195 of 605
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.

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196 Hard
196 of 605
Application +4 | -1
If sin⁡−1x+sin⁡−1y+sin⁡−1z=          3π2  and  f(1)=2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{\;\;\;\;\;3\pi }}{2}\;\mathrm{and}\;f\left( 1 \right) = 2,

f(p+q)=f(p).f(q),∀  p,  q∈R,  f\left( {p + q} \right) = f\left( p \right).f\left( q \right),\forall \;p,\;q \in R,\; then

xf(1)+yf(2)+zf(3)−(x+y+z)xf(1)+yf(2)+zf(3)  {x^{f\left( 1 \right)}} + {y^{f\left( 2 \right)}} + {z^{f\left( 3 \right)}} - \frac{{\left( {x + y + z} \right)}}{{{x^{f\left( 1 \right)}} + {y^{f\left( 2 \right)}} + {z^{f\left( 3 \right)}}}}\; is equal to
View answer and explanation
Correct option: C
2
Option A (2) is correct because given the conditions, x, y, and z must be 1, and using the functional equation, we can find f(1), f(2), and f(3), then substitute to get the result.
197 Medium
197 of 605
Application +4 | -1
If cos⁡−1p+cos⁡−1q+cos⁡−1r=3π{\cos ^{ - 1}}p + {\cos ^{ - 1}}q + {\cos ^{ - 1}}r = 3\pi, then p2+q2+r2+2pqr{p^2} + {q^2} + {r^2} + 2pqr is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because the given condition cos⁡−1p+cos⁡−1q+cos⁡−1r=3π{\cos ^{ - 1}}p + {\cos ^{ - 1}}q + {\cos ^{ - 1}}r = 3\pi implies p=q=r=−1p = q = r = -1, and thus p2+q2+r2+2pqr=1{p^2} + {q^2} + {r^2} + 2pqr = 1.
198 Easy
198 of 605
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.
199 Easy
199 of 605
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.
200 Medium
200 of 605
Application +4 | -1
The value of sin⁡−1(1213)−sin⁡−1(35)\sin ^{-1}\left(\frac{12}{13}\right)-\sin ^{-1}\left(\frac{3}{5}\right) is equal to :
View answer and explanation
Correct option: B
π2−sin⁡−1(5665)\frac{\pi}{2}-\sin ^{-1}\left(\frac{56}{65}\right)
The correct answer is B (π2−sin⁡−1(5665)\frac{\pi}{2}-\sin^{-1}(\frac{56}{65})) because it involves using the sine subtraction formula and trigonometric identities.
201 Easy
201 of 605
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.
202 Easy
202 of 605
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.
203 Easy
203 of 605
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.
204 Easy
204 of 605
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}.
205 Easy
205 of 605
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.
206 Medium
206 of 605
Application +4 | -1
The value of cos⁡−1[cot⁡{sin⁡−1(2−34)+cos⁡−1(124)+sec⁡−1(2)}]\cos ^{-1}\left[\cot \left\{\sin ^{-1}\left(\sqrt{\frac{2-\sqrt{3}}{4}}\right)+\cos ^{-1}\left(\frac{\sqrt{12}}{4}\right)+\sec ^{-1}(\sqrt{2})\right\}\right] is equal to
View answer and explanation
Correct option: A
π2\frac{\pi}{2}
Option A (π2\frac{{\pi}}{{2}}) is correct because by simplifying the expression inside the inverse cosine function, we find that it equals zero, and the inverse cosine of zero is π2\frac{{\pi}}{{2}}.
207 Easy
207 of 605
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.
208 Easy
208 of 605
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.
209 Easy
209 of 605
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.
210 Easy
210 of 605
Analyzing +4 | -1
If x,y and z are in A.P and tan⁡−1x,  tan⁡−1y    and   tan⁡−1z{\tan ^{ - 1}}x,\,\,{\tan ^{ - 1}}y\,\,\,\,\mathrm{and}\,\,\,{\tan ^{ - 1}}z are also in AP, then
View answer and explanation
Correct option: A
x=y=zx = y = z
Option A (x=y=zx = y = z) is correct because if x, y, z are in both A.P. and their arctangents are in A.P., it implies they must be equal.
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