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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Mixed difficulty

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Practice format and chapter quiz

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Showing 1-30 of 605 questions
1 Medium
1 of 605
Application +4 | -1
The value of cos⁡−1(cos⁡12)−sin⁡−1(sin⁡12)\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12) is-
View answer and explanation
Correct option: C
8π−48 \pi-4
Option C (8π−248 \pi - 24) is correct because it involves finding the correct range for the inverse trigonometric functions and then subtracting them.
2 Medium
2 of 605
Application +4 | -1
The value of sin⁡(2tan⁡−1(1/3))+\sin \left(2 \tan ^{-1}(1 / 3)\right)+ cos⁡(tan⁡−122)\cos \left(\tan ^{-1} 2 \sqrt{2}\right) is
View answer and explanation
Correct option: C
14/1514 / 15
Option C (14/15) is correct because by evaluating the sine and cosine terms using trigonometric identities and substituting, we arrive at the correct answer.
3 Medium
3 of 605
Application +4 | -1
Find cos⁡−1{12(cos⁡9π10−sin⁡9π10)}=\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}=
View answer and explanation
Correct option: D
17π20\frac{17 \pi}{20}
Option D (17π20\frac{{17 \pi}}{{20}}) is correct because it simplifies the given expression using trigonometric identities and inverse cosine properties.
4 Hard
4 of 605
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}.
5 Easy
5 of 605
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}.
6 Easy
6 of 605
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.
7 Medium
7 of 605
Application +4 | -1
The number of positive integral solutions of the equation tan⁡−1x+cot⁡−1y=tan⁡−14\tan ^{-1} x+\cot ^{-1} y=\tan ^{-1} 4 is
View answer and explanation
Correct option: B
1
Option B (1) is correct because by simplifying the equation, we find only one positive integral solution that satisfies the given condition.
8 Medium
8 of 605
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 the 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 correct because it is derived by using trigonometric identities to simplify the given equation and solve for xx.
9 Medium
9 of 605
Application +4 | -1
The value of α22cosec⁡2(12tan⁡−1αβ)−β22sec⁡2(12tan⁡−1βα)\frac{\alpha^2}{2} \operatorname{cosec}^2\left(\frac{1}{2} \tan ^{-1} \frac{\alpha}{\beta}\right)-\frac{\beta^2}{2} \sec ^2\left(\frac{1}{2} \tan ^{-1} \frac{\beta}{\alpha}\right) is equal to
View answer and explanation
Correct option: D
α2+β2(β+α)\sqrt{\alpha^2+\beta^2}(\beta+\alpha)
Option D (α2+β2(β+α)\sqrt{{\alpha^2+\beta^2}}(\beta+\alpha)) is correct because after simplifying the trigonometric expressions and substituting, the value equals to the given expression.
10 Medium
10 of 605
Application +4 | -1
2(tan⁡−11+tan⁡−12+tan⁡−13)2\left(\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3\right) is equal to
View answer and explanation
Correct option: D
2π2 \pi
Option D (2π2 \pi) is correct because the expression 2(tan⁡−11+tan⁡−12+tan⁡−13)2(\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3) simplifies to 2π2\pi using the properties of inverse tangent functions.
11 Easy
11 of 605
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}.
12 Easy
12 of 605
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.
13 Easy
13 of 605
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.
14 Medium
14 of 605
Application +4 | -1
sin⁡−1x5+cosec⁡−154=π2\sin ^{-1} \frac{x}{5}+\operatorname{cosec}^{-1} \frac{5}{4}=\frac{\pi}{2} then x=x=
View answer and explanation
Correct option: D
3
Option D (3) is correct because using the identity sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} and cosec⁡−1x=sin⁡−11x\operatorname{cosec}^{-1} x = \sin^{-1} \frac{1}{x}, we can solve for x.
15 Easy
15 of 605
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.

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16 Medium
16 of 605
Application +4 | -1
Two angles of a triangle are sin⁡−1(1/5)\sin ^{-1}(1 / \sqrt{5}) and sin⁡−1(1/10)\sin ^{-1}(1 / \sqrt{10}) then the third angle is
View answer and explanation
Correct option: B
3π/43 \pi / 4
Option B, 3π/43 \pi / 4, is correct because by calculating the sine of the sum of the two given angles, we can find the third angle of the triangle.
17 Easy
17 of 605
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.
18 Medium
18 of 605
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: A
14512\frac{\sqrt{145}}{12}
The correct answer is A (14512\frac{\sqrt{145}}{12}) because it involves using inverse cosine properties and solving the resulting equation.
19 Medium
19 of 605
Application +4 | -1
If in a triangle, two angles are tan⁡−12\tan ^{-1} 2 and tan⁡−13\tan ^{-1} 3, then the third angle is
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 greater than π2\frac{{\pi}}{{2}}, requiring the use of the adjusted tangent addition formula, leading to this third angle.
20 Easy
20 of 605
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.
21 Easy
21 of 605
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.
22 Easy
22 of 605
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.
23 Medium
23 of 605
Application +4 | -1
If 2 tan⁡−1(cos  x)=tan⁡−1(2  cosec  x),  {\tan ^{ - 1}}\left( {\mathrm{cos}\;x} \right) = {\tan ^{ - 1}}\left( {2\;\mathrm{cosec}\;x} \right),\; then the value of xx is
View answer and explanation
Correct option: B
π4\frac{\pi }{4}
Option B is correct because solving the equation 2tan⁡−1(cos⁡x)=tan⁡−1(2cosec  x)2{\tan ^{ - 1}}(\cos x) = {\tan ^{ - 1}}(2\mathrm{cosec}\;x) leads to x=π4x = \frac{\pi }{4}.
24 Easy
24 of 605
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}.
25 Medium
25 of 605
Application +4 | -1
If sin⁡−1(15)+sec⁡−1(2)+2tan⁡−1(13)+sec⁡−1(5)\sin ^{-1}\left(\frac{1}{5}\right)+\sec ^{-1}(2)+2 \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\sec ^{-1}(5) +sin⁡−1(12)+2tan⁡−1(3)=kπ+\sin ^{-1}\left(\frac{1}{2}\right)+2 \tan ^{-1}(\sqrt{3})={k} \pi, then k={k}=
View answer and explanation
Correct option: A
2
Option A (2) is correct because by using trigonometric identities and simplifying, the value of k is found to be 2.
26 Medium
26 of 605
Application +4 | -1
If tan⁡−1x+tan⁡−1y+tan⁡−1z=π\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi then 1xy+1yz+1zx\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} is
View answer and explanation
Correct option: B
1
Option B (1) is correct because if tan⁡−1x+tan⁡−1y+tan⁡−1z=π\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi, then x+y+z=xyzx + y + z = xyz, which implies 1xy+1yz+1zx=1\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = 1.
27 Medium
27 of 605
Application +4 | -1
If tan⁡−1x−tan⁡−1y=tan⁡−1A\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} A, and xy>−1x y>-1, then AA is
View answer and explanation
Correct option: C
x−y1+xy\frac{x-y}{1+x y}
Option C (x−y1+xy\frac{x-y}{1+xy}) is correct because using the tangent subtraction formula, tan⁡−1x−tan⁡−1y=tan⁡−1x−y1+xy\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}, which directly gives the value of A.
28 Easy
28 of 605
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.
29 Medium
29 of 605
Application +4 | -1
The value of cot⁡−11−sin⁡x+1+sin⁡x1−sin⁡x−1+sin⁡x\cot -1 \frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}} if 0∘<x<90∘0^{\circ} < x < 90^{\circ} is
View answer and explanation
Correct option: C
π−x2\pi-\frac{x}{2}
Option C (π−x2\pi - \frac{x}{2}) is correct because simplifying the expression inside the cot⁡−1\cot^{-1} function leads to −cot⁡(x2)-\cot(\frac{x}{2}), and using the property cot⁡−1(−x)=π−cot⁡−1(x)\cot^{-1}(-x) = \pi - \cot^{-1}(x) gives the result.
30 Medium
30 of 605
Application +4 | -1
If (tan⁡−1x)2+(cot⁡−1x)2=5π2/8\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=5 \pi^2 / 8, then xx equals
View answer and explanation
Correct option: C
-1
Option C (-1) is correct because substituting x = -1 into the equation satisfies the given condition.
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Key Takeaways

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