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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Easy

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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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Showing 1-30 of 365 questions
1 Easy
1 of 365
Application +4 | -1
If x2+y2+z2=r2,{x^2} + {y^2} + {z^2} = {r^2}, then

tan⁡−1(xyzr)+tan⁡−1(yzxr)+tan⁡−1(xzyr)  {\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\; is equal to
View answer and explanation
Correct option: B
π2\frac{\pi }{2}
Option B (π2\frac{\pi }{2}) is correct because the expression simplifies to tan⁡−1(∞)\tan^{-1}(\infty), which equals π2\frac{\pi }{2} given the condition x2+y2+z2=r2{x^2} + {y^2} + {z^2} = {r^2}.
2 Easy
2 of 365
Application +4 | -1
2tan⁡−1(13)+tan⁡−1(17)2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) is equal to
View answer and explanation
Correct option: D
π4\frac{\pi }{4}
Option D (π4\frac{\pi }{4}) is correct because it is the result of applying the tangent addition formula twice to simplify the expression.
3 Easy
3 of 365
Application +4 | -1
If −1≤x≤−12- 1 \le x \le - \frac{1}{2}, then sin⁡−1(3x−4x3){\sin ^{ - 1}}\left( {3x - 4{x^3}} \right) equals
View answer and explanation
Correct option: C
−π−3sin⁡−1x- \pi - 3{\sin ^{ - 1}}x
Option D (3\sqrt 3) is correct because the expression simplifies to tan⁡(2π5−π15)=tan⁡(π3)=3\tan(\frac{2\pi}{5} - \frac{\pi}{15}) = \tan(\frac{\pi}{3}) = \sqrt{3}.
4 Easy
4 of 365
Analyzing +4 | -1
For the equation cos⁡−1x+cos⁡−12x+π=0{\cos ^{ - 1}}x + {\cos ^{ - 1}}2x + \pi = 0, then the number of real solutions is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the equation has no real solutions, as the value of x obtained does not satisfy the original equation.
5 Easy
5 of 365
Application +4 | -1
If sin⁡−1a+sin⁡−1b+sin⁡−1c=π,{\sin ^{ - 1}}a + {\sin ^{ - 1}}b + {\sin ^{ - 1}}c = \pi , then the value of

a(1−a2)+b(1−b2)+c1−c2)  a\sqrt {\left( {1 - {a^2}} \right)} + b\sqrt {\left( {1 - {b^2}} \right)} + c\sqrt {1 - {c^2})} \; will be
View answer and explanation
Correct option: A
2 abcabc
Option A (2 abcabc) is correct because it is derived from the trigonometric identity sin⁡2A+sin⁡2B+sin⁡2C=4sin⁡Asin⁡Bsin⁡C\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C when A+B+C=πA + B + C = \pi, where A=sin⁡−1aA = {\sin ^{ - 1}}a, B=sin⁡−1bB = {\sin ^{ - 1}}b, and C=sin⁡−1cC = {\sin ^{ - 1}}c.
6 Easy
6 of 365
Application +4 | -1
If tan⁡−11+x2−1x=4\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x}=4 then x=x=
View answer and explanation
Correct option: D
tan⁡8\tan 8
Option D (tan⁡8\tan 8) is correct because substituting x=tan⁡θx = \tan \theta simplifies the expression to (1/2)tan⁡−1(x)=4(1/2) \tan^{-1}(x) = 4, leading to x=tan⁡8x = \tan 8.
7 Easy
7 of 365
Application +4 | -1
The value of tan⁡(2tan⁡−1(35)+sin⁡−1(513))\tan \left( {2{{\tan }^{ - 1}}\left( {\frac{3}{5}} \right) + {{\sin }^{ - 1}}\left( {\frac{5}{{13}}} \right)} \right) is equal to
View answer and explanation
Correct option: B
22021\frac{{220}}{{21}}
Option B (220/21) is correct because using trigonometric identities and tangent addition formula, the expression simplifies to 220/21.
8 Easy
8 of 365
Application +4 | -1
The value of tan {cos⁡−1(−27)−π2}\left\{ {{{\cos }^{ - 1}}\left( { - \frac{2}{7}} \right) - \frac{\pi }{2}} \right\} is
View answer and explanation
Correct option: A
235\frac{2}{{3\sqrt 5 }}
Option A (235\frac{2}{{3\sqrt 5 }}) is correct because after applying trigonometric identities and simplifications, the expression evaluates to this value.
9 Easy
9 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 D (a−b1+ab\frac{{a - b}}{{1 + ab}}) is the correct answer because it is derived from applying the tangent subtraction formula after converting inverse sine and cosine to inverse tangent.
10 Easy
10 of 365
Application +4 | -1
Let cos⁡(2tan⁡−1x)=12\cos \left( {2{{\tan }^{ - 1}}x} \right) = \frac{1}{2}, then the value of xx is
View answer and explanation
Correct option: B
13\frac{1}{{\sqrt 3 }}
Option B (13\frac{1}{{\sqrt 3 }}) is correct because it involves solving the equation using the inverse trigonometric function and the cosine function.
11 Easy
11 of 365
Application +4 | -1
If ∑i=120sin⁡−1xi=10π,\sum \limits_{i = 1}^{20} {\sin ^{ - 1}}{x_i} = 10\pi , then ∑i=120xi\sum \limits_{i = 1}^{20} {x_i} is equal to
View answer and explanation
Correct option: A
20
Option A (20) is correct because the maximum value of sin⁡−1xi{\sin ^{ - 1}}{x_i} is π2\frac{\pi }{2}, and for the sum to be 10π10\pi, each term must be π2\frac{\pi }{2}.
12 Easy
12 of 365
Application +4 | -1
The value of cos⁡(3cos⁡−145)+sin⁡(3sin⁡−145)\cos \left(3 \cos ^{-1} \frac{4}{5}\right)+\sin \left(3 \sin ^{-1} \frac{4}{5}\right) is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because by applying the triple angle formulas for cosine and sine and simplifying, the expression evaluates to zero.
13 Easy
13 of 365
Analyzing +4 | -1
The maximum value of sec⁡−1(7−5(x2+3)2(x2+2))\sec ^{-1}\left(\frac{7-5\left(x^{2}+3\right)}{2\left(x^{2}+2\right)}\right) is:
View answer and explanation
Correct option: D
2π3\frac{2 \pi}{3}
Option D (2π/3) is correct because it represents the maximum value of the inverse secant function after simplifying the expression inside.
14 Easy
14 of 365
Analyzing +4 | -1
If x=sin⁡−1(sin⁡10)x=\sin ^{-1}(\sin 10) and y=cos⁡−1(cos⁡10)y=\cos ^{-1}(\cos 10), then y−xy-x is equal to:
View answer and explanation
Correct option: D
π\pi
Option D (π\pi) is correct because, after finding the correct ranges for the inverse trigonometric functions, the difference y-x simplifies to π\pi.
15 Easy
15 of 365
Application +4 | -1
If cos⁡−1x+cos⁡−1(y/2)=α\cos ^{-1} x+\cos ^{-1}(y / 2)=\alpha, then 4x2−4xycos⁡α+y24 x^2-4 x y \cos \alpha+y^2 is equal to
View answer and explanation
Correct option: A
4sin⁡2α4 \sin ^2 \alpha
Option A (4sin⁡2α4 \sin ^2 \alpha) is correct because by applying the cosine subtraction formula and simplifying, the expression 4x2−4xycos⁡α+y24 x^2-4 x y \cos \alpha+y^2 is equal to 4sin⁡2α4 \sin ^2 \alpha.

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16 Easy
16 of 365
Analyzing +4 | -1
If sin⁡−1x+sin⁡−1y+sin⁡−1z=π,  then  x4+y4+z4+4x2y2z2=k(x2y2+y2z2+z2x2){\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi ,\;\mathrm{then}\;{x^4} + {y^4} + {z^4} + 4{x^2}{y^2}{z^2} = k\left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2}} \right) Where kk is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because after applying trigonometric identities and simplifying the given equation, the value of k is found to be 2.
17 Easy
17 of 365
Application +4 | -1
If 0≤x<∞0 \le x < \infty, then cos⁡−1(1−x21+x2){\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) equals
View answer and explanation
Correct option: A
2tan⁡−1x2{\tan ^{ - 1}}x
Option A (2tan⁡−1x2{\tan ^{ - 1}}x) is correct because it directly applies the trigonometric identity cos⁡−1(1−x21+x2)=2tan⁡−1x{\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) = 2{\tan ^{ - 1}}x for 0≤x<∞0 \le x < \infty.
18 Easy
18 of 365
Application +4 | -1
The solution of tan⁡−12x+tan⁡−13x=π4{\tan ^{ - 1}}2x + {\tan ^{ - 1}}3x = \frac{\pi }{4} is
View answer and explanation
Correct option: A
16\frac{1}{6}
Option A (16\frac{1}{6}) is correct because it is the only solution that satisfies the original equation after checking for extraneous roots.
19 Easy
19 of 365
Application +4 | -1
tan⁡−1c1x−yc1y+x+tan⁡−1c2−c11+c2c1+tan⁡−1c3−c21+c3c2+...+tan⁡−11cn{\tan ^{ - 1}}\frac{{{c_1}x - y}}{{{c_1}y + x}} + {\tan ^{ - 1}}\frac{{{c_2} - {c_1}}}{{1 + {c_2}{c_1}}} + {\tan ^{ - 1}}\frac{{{c_3} - {c_2}}}{{1 + {c_3}{c_2}}} + ... + {\tan ^{ - 1}}\frac{1}{{{c_n}}} is equal to
View answer and explanation
Correct option: C
tan⁡−1xy{\tan ^{ - 1}}\frac{x}{y}
Option C is correct because the given expression simplifies to tan⁡−1xy{\tan ^{ - 1}}\frac{x}{y} due to the telescoping nature of the series.
20 Easy
20 of 365
Application +4 | -1
The value of cot⁡(∑n=119cot⁡−1(1+∑p=1n2p))\cot \left(\sum_{n=1}^{19} \cot ^{-1}\left(1+\sum_{p=1}^n 2 p\right)\right) is
View answer and explanation
Correct option: D
2119\frac{21}{19}
Option D (2119\frac{21}{19}) is correct because the summation telescopes, simplifying the expression inside the cotangent function.
21 Easy
21 of 365
Analyzing +4 | -1
The number of solutions of the equation, sin⁡−1x=2tan⁡−1x\sin ^{-1} x=2 \tan ^{-1} x (in principal values) is :
View answer and explanation
Correct option: A
1
Option A (1) is correct because the equation sin⁡−1x=2tan⁡−1x\sin^{-1}x = 2 \tan^{-1}x has only one solution, which is x=0.
22 Easy
22 of 365
Understanding +4 | -1
cos⁡(cos⁡−1(725))\cos \left(\cos ^{-1}\left(\frac{7}{25}\right)\right) is equal to
View answer and explanation
Correct option: D
None of these
None of these is correct. The correct answer is 725\frac{7}{25} because cos⁡(cos⁡−1(x))=x\cos(\cos^{-1}(x)) = x for xx in the domain of cos⁡−1(x)\cos^{-1}(x), which is [−1,1][-1, 1].
23 Easy
23 of 365
Application +4 | -1
A value of xx satisfying tan⁡−1(x+3)−tan⁡−1(x−3)=sin⁡−1(3/5)\tan ^{-1}(x+3)-\tan ^{-1}(x-3)=\sin ^{-1}(3 / 5) is
View answer and explanation
Correct option: C
4
Option C (4) is the correct answer because it satisfies the given equation after applying the tangent subtraction formula and solving the resulting quadratic equation.
24 Easy
24 of 365
Analyzing +4 | -1
If α=sin⁡−132+sin⁡−113\alpha = {\sin ^{ - 1}}\frac{{\sqrt 3 }}{2} + {\sin ^{ - 1}}\frac{1}{3} and β=cos⁡−132+cos⁡−113\beta = {\cos ^{ - 1}}\frac{{\sqrt 3 }}{2} + {\cos ^{ - 1}}\frac{1}{3}, then
View answer and explanation
Correct option: C
α<β\alpha < \beta
Option C (α<β\alpha < \beta) is correct because by adding the two equations, we find α+β=π\alpha + \beta = \pi, and since α<π2\alpha < \frac{\pi}{2}, it follows that α<β\alpha < \beta.
25 Easy
25 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 equation tan⁡−1(ax)+tan⁡−1(bx)=π2{\tan ^{ - 1}}\left( {\frac{a}{x}} \right) + {\tan ^{ - 1}}\left( {\frac{b}{x}} \right) = \frac{\pi }{2} leads to x=abx = \sqrt{ab}.
26 Easy
26 of 365
Application +4 | -1
If 4cos⁡−1x+sin⁡−1x=π4 \cos ^{-1} x+\sin ^{-1} x=\pi, then xx equals to
View answer and explanation
Correct option: D
32\frac{\sqrt{3}}{2}
Option D (32\frac{\sqrt{3}}{2}) is correct because using the identity sin⁻¹x + cos⁻¹x = π/2 simplifies the equation, leading to x = √3/2.
27 Easy
27 of 365
Application +4 | -1
If tan⁡−11+x2−1−x21+x2+1−x2=α\tan ^{-1} \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}=\alpha, then x2x^2 is equal to
View answer and explanation
Correct option: D
sin⁡2α\sin 2 \alpha
Option D (sin⁡2α\sin 2 \alpha) is correct because after applying componendo and dividendo and simplifying, x2x^2 equals sin⁡2α\sin 2 \alpha.
28 Easy
28 of 365
Application +4 | -1
If 12≤x≤1\frac{1}{{\sqrt 2 }} \le x \le 1, then sin⁡−1(2x1−x2){\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) equals
View answer and explanation
Correct option: B
π−2sin⁡−1x\pi - 2{\sin ^{ - 1}}x
Option B (π−2sin⁡−1x\pi - 2{\sin ^{ - 1}}x) is correct because it accounts for the range of the inverse sine function when 12≤x≤1\frac{1}{{\sqrt 2 }} \le x \le 1.
29 Easy
29 of 365
Application +4 | -1
sin⁡−1(cos⁡(sin⁡−1x))+cos⁡−1(sin⁡(cos⁡−1x))\sin ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)+\cos ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right) is equal to
View answer and explanation
Correct option: B
π2\frac{\pi}{2}
Option B (π2\frac{{\pi}}{{2}}) is correct because using trigonometric identities simplifies the expression to π2\frac{{\pi}}{{2}}.
30 Easy
30 of 365
Application +4 | -1
Let cos⁡(2tan⁡−1x)=12\cos \left(2 \tan ^{-1} x\right)=\frac{1}{2}, then the value of xx will be
View answer and explanation
Correct option: B
13\frac{1}{\sqrt{3}}
Option B (13\frac{{1}}{{\sqrt{3}}}) is correct because it is the value of x that satisfies the given equation.
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Key Takeaways

  • ✓ This set contains 365 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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