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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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1 Hard
1 of 19
Application +4 | -1
If sin⁡(sin⁡−115+cos⁡−1x)=1\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1, then xx is equal to
View answer and explanation
Correct option: D
15\frac{1}{5}
Option D (15\frac{1}{5}) is correct because solving the equation sin⁡(sin⁡−115+cos⁡−1x)=1\sin(\sin^{-1} \frac{1}{5} + \cos^{-1} x) = 1 leads to x=15x = \frac{1}{5}.
2 Hard
2 of 19
Application +4 | -1
If ∑i=110cos⁡−1xi=0\sum\limits_{{i=1}}^{10} \cos ^{-1} x_i=0, then ∑i=110xi\sum\limits_{{i=1}}^{10} x_i is
View answer and explanation
Correct option: B
10
Option B (10) is correct because each term in the summation must be zero, implying each xix_i is 1, and the sum of ten 1's is 10.
3 Hard
3 of 19
Analyzing +4 | -1
If θ  \mathrm{\theta }\; and ϕ\phi are the roots of the equation 8x2+22x+5=0,8{x^2} + 22x + 5 = 0, then
View answer and explanation
Correct option: C
Both tan⁡−1θ{\tan ^{ - 1}}\mathrm{\theta } and tan⁡−1ϕ{\tan ^{ - 1}}\phi are equal
Option C (Both tan⁡−1θ{\tan ^{ - 1}}\theta and tan⁡−1ϕ{\tan ^{ - 1}}\phi are equal) is correct because both roots of the quadratic equation are real numbers, and the arctangent function is defined for all real numbers.
4 Hard
4 of 19
Analyzing +4 | -1
Let s={x∈R:0<x<1s=\left\{x \in \mathrm{R}: 0 < x < 1\right. and 2tan⁡−1(1−x1+x)=cos⁡−1(1−x21+x2)}\left.2 \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right\} If n(S)n(S) denotes the number of elements in SS then:
View answer and explanation
Correct option: C
n(S)=1n(S)=1 and the element in SS is less than 12\frac{1}{2}
Option C is correct because solving the given equation using trigonometric identities yields a single solution for x, which is less than 1/2.
5 Hard
5 of 19
Application +4 | -1
If f(x)=sin⁡−1{32x−121−x2},−72≤x≤1f(x)=\sin ^{-1}\left\{\frac{\sqrt{3}}{2} x-\frac{1}{2} \sqrt{1-x^2}\right\},-\frac{7}{2} \leq x \leq 1, then f(x)f(x) is equal to f(x)f(x) is equal to
View answer and explanation
Correct option: B
sin⁡−1x−π6\sin ^{-1} x-\frac{\pi}{6}
Option B (sin⁡−1x−π6\sin ^{-1} x-\frac{\pi}{6}) is correct because the given expression simplifies to sin⁡−1x−π6\sin^{-1} x - \frac{\pi}{6} using trigonometric identities and substitution.
6 Hard
6 of 19
Application +4 | -1
The value tan⁡(12cos⁡−153)\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right) is
View answer and explanation
Correct option: B
3−52\frac{3-\sqrt{5}}{2}
Option B ((3-√5)/2) is correct because using the half-angle formula for tangent and simplifying leads to the correct result.
7 Hard
7 of 19
Application +4 | -1
If tan⁡−1x+cos⁡−1y(1+y2)=sin⁡−1310\tan ^{-1} x+\cos ^{-1} \frac{y}{\sqrt{\left(1+y^2\right)}}=\sin ^{-1} \frac{3}{\sqrt{10}} and both xx and yy are positive and integral, then xx and yy is equal to
View answer and explanation
Correct option: A
(1,2)(1,2) and (2,7)(2,7)
Option A ((1,2) and (2,7)) is correct because solving the given equation with the constraints that x and y are positive integers leads to these two solution pairs.
8 Hard
8 of 19
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.
9 Hard
9 of 19
Analyzing +4 | -1
If sin⁡−1α17+cos⁡−145−tan⁡−17736=0,0<α<13\sin ^{-1} \frac{\alpha}{17}+\cos ^{-1} \frac{4}{5}-\tan ^{-1} \frac{77}{36}=0,0<\alpha<13, then sin⁡−1(sin⁡α)+cos⁡−1(cos⁡α)\sin ^{-1}(\sin \alpha)+\cos ^{-1}(\cos \alpha) is equal to
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because by solving the given equation for α\alpha, we find α=8\alpha = 8, and then sin⁡−1(sin⁡8)+cos⁡−1(cos⁡8)\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8) simplifies to π\pi.
10 Hard
10 of 19
Analyzing +4 | -1
Statement I: The equation (sin⁡−1x)3+(cos⁡−1x)3−aπ3=0\left(\sin ^{-1} \mathrm{x}\right)^{3}+\left(\cos ^{-1} \mathrm{x}\right)^{3}-\mathrm{a} \pi^{3}=0 has a solution for all a≥132\mathrm{a} \geq \frac{1}{32}.

Statement II: For any x∈R,sin⁡−1x+cos⁡−1x=π2x \in R, \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} and 0≤(sin⁡−1x−π4)2≤9π2160 \leq\left(\sin ^{-1} \mathrm{x}-\frac{\pi}{4}\right)^{2} \leq \frac{9 \pi^{2}}{16}
View answer and explanation
Correct option: A
Both statements I and II are true.
Statement I and II are both true because statement II correctly states the relationship between inverse sine and cosine, and statement I is true because the given equation has a solution for a within the specified range.
11 Hard
11 of 19
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 x100+y100+z100−9x101+y101+z101{x^{100}} + {y^{100}} + {z^{100}} - \frac{9}{{{x^{101}} + {y^{101}} + {z^{101}}}} is
View answer and explanation
Correct option: A
0
Option D (0,  ±120,\; \pm \frac{1}{2}) is correct because solving the equation using the tangent addition formula leads to these three solutions.
12 Hard
12 of 19
Analyzing +4 | -1
The value of ∑r=0∞tan⁡−1(11+r+r2)  \sum \limits_{r = 0}^\infty {\tan ^{ - 1}}\left( {\frac{1}{{1 + r + {r^2}}}} \right)\; is equal to
View answer and explanation
Correct option: A
π2\frac{\pi }{2}
Option A (π2\frac{\pi }{2}) is correct because the sum telescopes to tan⁡−1(∞)\tan^{-1}(\infty), which equals π2\frac{\pi }{2}.
13 Hard
13 of 19
Application +4 | -1
If tan⁡−12x+tan⁡−13x=π4\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}, then xx is equal to
View answer and explanation
Correct option: B
16\frac{1}{6}
Option B (16\frac{1}{6}) is correct because solving the equation tan⁡−12x+tan⁡−13x=π4\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4} leads to x=16x = \frac{1}{6}, after discarding the extraneous solution.
14 Hard
14 of 19
Understanding +4 | -1
cos⁡−1a−xa−b=sin⁡−1x−ba−b\cos ^{-1} \sqrt{\frac{a-x}{a-b}}=\sin ^{-1} \sqrt{\frac{x-b}{a-b}} is possible if
View answer and explanation
Correct option: A
a>x>ba > x > b
Option A (a>x>ba > x > b) is correct because it satisfies the conditions for the inverse trigonometric functions to be defined and equal.
15 Hard
15 of 19
Application +4 | -1
If sin⁡−1(−x22+x34−…)+cos⁡−1(x2−x42+x64−…)\sin ^{-1}\left(-\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) =π2=\frac{\pi}{2} for 0<∣x∣<20<|x|<\sqrt{2}, then xx equals
View answer and explanation
Correct option: B
1
Option B (1) is correct because simplifying the given equation using the properties of inverse trigonometric functions and geometric series leads to x = 1.

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16 Hard
16 of 19
Application +4 | -1
If the solution set of system of equations 3sin⁡−1x=−π3 \sin ^{-1} x=-\pi −sin⁡−1(3x−4x3)-\sin ^{-1}\left(3 x-4 x^3\right) is [a,b][a, b] and that of 3sin⁡−1x=π−sin⁡−13 \sin ^{-1} x=\pi-\sin ^{-1} (3x−4x3)\left(3 x-4 x^3\right) is [c,d][\mathrm{c}, \mathrm{d}] then evaluate ∣a∣+∣b∣+∣c∣+∣d∣|a|+|b|+|c|+|d|
View answer and explanation
Correct option: A
3
Option A (3) is correct because the solution sets are [-1, -1/2] and [1/2, 1], so |a| + |b| + |c| + |d| = 3.
17 Hard
17 of 19
Application +4 | -1
tan⁡[12cos⁡−1(53)]\tan \left[\frac{1}{2} \cos ^{-1}\left(\frac{\sqrt{5}}{3}\right)\right] is equal to
View answer and explanation
Correct option: A
3−52\frac{3-\sqrt{5}}{2}
Option A (3−52\frac{3-\sqrt{5}}{2}) is correct because using the half-angle formula for tangent and simplifying the expression leads to the correct answer.
18 Hard
18 of 19
Understanding +4 | -1
sin⁡−1x+sin⁡−11x+cos⁡−1x+cos⁡−11x{\sin ^{ - 1}}x + {\sin ^{ - 1}}\frac{1}{x} + {\cos ^{ - 1}}x + {\cos ^{ - 1}}\frac{1}{x} is equal to
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because the sum of inverse trigonometric functions simplifies to π\pi using the properties of inverse sine and cosine.
19 Hard
19 of 19
Analyzing +4 | -1
Domain of the function f(x)=2Sin⁡−11−x+Sin⁡−1[2x(1−x)]f(x)=2 \operatorname{Sin}^{-1} \sqrt{1-x}+\operatorname{Sin}^{-1}[2 \sqrt{x(1-x)}] is
View answer and explanation
Correct option: A
0≤x≤10 \leq x \leq 1
Option A (0≤x≤10 \leq x \leq 1) is correct because it satisfies the domain restrictions of both inverse sine functions, ensuring that the arguments are within [-1, 1].
Revision notes and key takeaways

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Key Takeaways

  • ✓ This set contains 19 single-correct-answer MCQs on INVERSE TRIGONOMETRY.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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