Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Mathematics PYQs
  3. /Inverse Trigonometric Functions

Inverse Trigonometric Functions — JEE Previous Year Questions

Every Inverse Trigonometric Functions question asked in JEE Main and JEE Advanced across the last 186 papers — 101 questions, each with its correct answer. Free to read, no account needed.

Questions

101

Papers it appeared in

93/186

Appearance rate

50%

All 101 Inverse Trigonometric Functions questions

Most recent papers first.

Q1·MathematicsSingle correctJEE Advanced 2026
Considering only the principal values of the inverse trigonometric functions, the value of cot⁡−1(cot⁡(−11))+10sin⁡(2cos⁡−1(12))+10sin⁡(2tan⁡−1(2))\cot^{-1}(\cot(-11)) + 10 \sin\left( 2 \cos^{-1}\left( \frac{1}{\sqrt{2}} \right) \right) + 10 \sin(2 \tan^{-1}(2))cot−1(cot(−11))+10sin(2cos−1(2​1​))+10sin(2tan−1(2)) is
  1. (A)3π+73\pi + 73π+7
  2. (B)777
  3. (C)4π+74\pi + 74π+7
  4. (D)3π−53\pi - 53π−5

Correct answer: (C)

Step-by-step solution →
Q2·MathematicsSingle correctJEE Main 2026
Let α=3sin⁡−1(611)\alpha = 3\sin^{-1}\left(\frac{6}{11}\right)α=3sin−1(116​) and β=3cos⁡−1(49)\beta = 3\cos^{-1}\left(\frac{4}{9}\right)β=3cos−1(94​), where inverse trigonometric functions take only the principal values. Given below are two statements: Statement I: cos⁡(α+β)>0\cos(\alpha + \beta) > 0cos(α+β)>0. Statement II: cos⁡(α)<0\cos(\alpha) < 0cos(α)<0. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both Statement I and Statement II are true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (A)

Step-by-step solution →
Q3·MathematicsSingle correctJEE Main 2026
Let 0 < α < 1, β=13α\beta = \frac{1}{3\alpha}β=3α1​ and tan−1^{-1}−1(1 − α) + tan−1^{-1}−1(1 − β) = π4\frac{\pi}{4}4π​. Then 6(α + β) is equal to:
  1. (A)6
  2. (B)7
  3. (C)8
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
If sin⁡(tan⁡−1(x2))=cot⁡(sin⁡−11−x2)\sin(\tan^{-1}(x\sqrt{2})) = \cot(\sin^{-1}\sqrt{1 - x^{2}})sin(tan−1(x2​))=cot(sin−11−x2​), x∈(0,1)x \in (0, 1)x∈(0,1), then the value of xxx is :
  1. (A)12\frac{1}{2}21​
  2. (B)13\frac{1}{3}31​
  3. (C)23\frac{2}{3}32​
  4. (D)58\frac{5}{8}85​

Correct answer: (A)

Step-by-step solution →
Q5·MathematicsNumericalJEE Main 2026
If π4+∑p=111tan⁡−1(2p−11+22p−1)=α\frac{\pi}{4} + \sum\limits_{p=1}^{11} \tan^{-1}\left(\frac{2^{p-1}}{1 + 2^{2p-1}}\right) = \alpha4π​+p=1∑11​tan−1(1+22p−12p−1​)=α, then tan⁡α\tan\alphatanα is equal to __________.

Correct answer: 2048

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
Let [⋅][\cdot][⋅] denote the greatest integer function. If the domain of the function f(x)=cos⁡−1(4x+2[x]3)f(x) = \cos^{-1}\left(\frac{4x + 2[x]}{3}\right)f(x)=cos−1(34x+2[x]​) is [α,β][\alpha, \beta][α,β], then 12(α+β)12(\alpha + \beta)12(α+β) is equal to:
  1. (A)666
  2. (B)888
  3. (C)999
  4. (D)444

Correct answer: (A)

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
If y=tan⁡−1(3cos⁡x−4sin⁡x4cos⁡x+3sin⁡x)+2tan⁡−1(x1+1−x2)y = \tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right) + 2\tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)y=tan−1(4cosx+3sinx3cosx−4sinx​)+2tan−1(1+1−x2​x​), then dydx\frac{dy}{dx}dxdy​ at x=32x = \frac{\sqrt{3}}{2}x=23​​ is equal to:
  1. (A)333
  2. (B)−1-1−1
  3. (C)111
  4. (D)222

Correct answer: (C)

Step-by-step solution →
Q8·MathematicsSingle correctJEE Main 2026
Considering the principal values of inverse trigonometric functions, the value of the expression tan⁡(2sin⁡−1(213)−2cos⁡−1(310))\tan\left(2\sin^{-1}\left(\dfrac{2}{\sqrt{13}}\right) - 2\cos^{-1}\left(\dfrac{3}{\sqrt{10}}\right)\right)tan(2sin−1(13​2​)−2cos−1(10​3​)) is equal to :
  1. (A)−3356-\dfrac{33}{56}−5633​
  2. (B)3356\dfrac{33}{56}5633​
  3. (C)1663\dfrac{16}{63}6316​
  4. (D)−1663-\dfrac{16}{63}−6316​

Correct answer: (B)

Step-by-step solution →
Q9·MathematicsNumericalJEE Main 2026
If k=tan⁡(π4+12cos⁡−1(23))+tan⁡(12sin⁡−1(23))k = \tan\left(\dfrac{\pi}{4} + \dfrac{1}{2}\cos^{-1}\left(\dfrac{2}{3}\right)\right) + \tan\left(\dfrac{1}{2}\sin^{-1}\left(\dfrac{2}{3}\right)\right)k=tan(4π​+21​cos−1(32​))+tan(21​sin−1(32​)) then the number of solutions of the equation sin⁡−1(kx−1)=sin⁡−1x−cos⁡−1x\sin^{-1}(kx - 1) = \sin^{-1}x - \cos^{-1}xsin−1(kx−1)=sin−1x−cos−1x is ______.

Correct answer: 1

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
The number of solutions of tan⁡−14x+tan⁡−16x=π6\tan^{-1} 4x + \tan^{-1}6x = \frac{\pi}{6}tan−14x+tan−16x=6π​, where −126<x<126-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}−26​1​<x<26​1​ is equal to
  1. (A)3
  2. (B)0
  3. (C)1
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
If the domain of the function f(x)=cos⁡−1(2x−511−3x)+sin⁡−1(2x2−3x+1)f(x) = \cos^{-1}\left(\frac{2x-5}{11-3x}\right) + \sin^{-1}(2x^{2} - 3x + 1)f(x)=cos−1(11−3x2x−5​)+sin−1(2x2−3x+1) is the interval [α,β][\alpha, \beta][α,β], then α+2β\alpha + 2\betaα+2β is equal to :
  1. (A)1
  2. (B)3
  3. (C)5
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q12·MathematicsNumericalJEE Main 2026
Let the maximum value of (sin⁡−1x)2+(cos⁡−1x)2(\sin^{-1}x)^{2} + (\cos^{-1}x)^{2}(sin−1x)2+(cos−1x)2 for x∈[−32,12]x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]x∈[−23​​,2​1​] be mnπ2\frac{m}{n}\pi^{2}nm​π2, where gcd (m,n)=1(m, n) = 1(m,n)=1. Then m+nm + nm+n is equal to _______.

Correct answer: 65

Step-by-step solution →
Q13·MathematicsSingle correctJEE Advanced 2025
The total number of real solutions of the equation θ=tan⁡−1(2tan⁡θ)−12sin⁡−1(6tan⁡θ9+tan⁡2θ)\theta = \tan^{-1}\left(2\tan\theta\right) - \frac{1}{2}\sin^{-1}\left( \frac{6\tan\theta}{9 + \tan^2\theta} \right)θ=tan−1(2tanθ)−21​sin−1(9+tan2θ6tanθ​) is (Here, the inverse trigonometric functions sin⁡−1x\sin^{-1}xsin−1x and tan⁡−1x\tan^{-1}xtan−1x assume values in [−π2,π2]\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right][−2π​,2π​] and (−π2,π2)\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)(−2π​,2π​), respectively.)
  1. (A)1
  2. (B)2
  3. (C)3
  4. (D)5

Correct answer: (C)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2025
The value of cot⁡−1(1+tan⁡2(2)−1tan⁡(2))−cot⁡−1(1+tan⁡2(12)+1tan⁡(12))\cot^{-1}\left(\dfrac{\sqrt{1+\tan^2(2)}-1}{\tan(2)}\right)-\cot^{-1}\left(\dfrac{\sqrt{1+\tan^2\left(\tfrac12\right)}+1}{\tan\left(\tfrac12\right)}\right)cot−1(tan(2)1+tan2(2)​−1​)−cot−1​tan(21​)1+tan2(21​)​+1​​ is equal to:
  1. (A)π−54\pi-\dfrac{5}{4}π−45​
  2. (B)π−32\pi-\dfrac{3}{2}π−23​
  3. (C)π+32\pi+\dfrac{3}{2}π+23​
  4. (D)π+52\pi+\dfrac{5}{2}π+25​

Correct answer: (A)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2025
The sum of the infinite series cot⁡−1(74)+cot⁡−1(194)+cot⁡−1(394)+cot⁡−1(674)+…\cot^{-1}\left(\dfrac{7}{4}\right)+\cot^{-1}\left(\dfrac{19}{4}\right)+\cot^{-1}\left(\dfrac{39}{4}\right)+\cot^{-1}\left(\dfrac{67}{4}\right)+\dotscot−1(47​)+cot−1(419​)+cot−1(439​)+cot−1(467​)+… is:
  1. (A)π2+tan⁡−1(12)\dfrac{\pi}{2}+\tan^{-1}\left(\dfrac{1}{2}\right)2π​+tan−1(21​)
  2. (B)π2−cot⁡−1(12)\dfrac{\pi}{2}-\cot^{-1}\left(\dfrac{1}{2}\right)2π​−cot−1(21​)
  3. (C)π2+cot⁡−1(12)\dfrac{\pi}{2}+\cot^{-1}\left(\dfrac{1}{2}\right)2π​+cot−1(21​)
  4. (D)π2−tan⁡−1(12)\dfrac{\pi}{2}-\tan^{-1}\left(\dfrac{1}{2}\right)2π​−tan−1(21​)

Correct answer: (D)

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2025
Considering the principal values of the inverse trigonometric functions, sin⁡−1(32x+121−x2)\sin^{-1}\left(\dfrac{\sqrt{3}}{2}x+\dfrac{1}{2}\sqrt{1-x^2}\right)sin−1(23​​x+21​1−x2​), −12<x<12-\dfrac{1}{2}<x<\dfrac{1}{\sqrt{2}}−21​<x<2​1​, is equal to:
  1. (A)π4+sin⁡−1x\dfrac{\pi}{4}+\sin^{-1}x4π​+sin−1x
  2. (B)π6+sin⁡−1x\dfrac{\pi}{6}+\sin^{-1}x6π​+sin−1x
  3. (C)−5π6−sin⁡−1x-\dfrac{5\pi}{6}-\sin^{-1}x−65π​−sin−1x
  4. (D)5π6−sin⁡−1x\dfrac{5\pi}{6}-\sin^{-1}x65π​−sin−1x

Correct answer: (B)

Step-by-step solution →
Q17·MathematicsIntegerJEE Main 2025
If y=cos⁡(π3+cos⁡−1x2)y=\cos\left(\dfrac{\pi}{3}+\cos^{-1}\dfrac{x}{2}\right)y=cos(3π​+cos−12x​), then (x−y)2+3y2(x-y)^2+3y^2(x−y)2+3y2 is equal to ______.

Correct answer: 3

Step-by-step solution →
Q18·MathematicsIntegerJEE Main 2025
Let S={x:cos⁡−1x=π+sin⁡−1x+sin⁡−1(2x+1)}S=\{x:\cos^{-1}x=\pi+\sin^{-1}x+\sin^{-1}(2x+1)\}S={x:cos−1x=π+sin−1x+sin−1(2x+1)}. Then ∑x∈S(2x−1)2\displaystyle\sum_{x\in S}(2x-1)^2x∈S∑​(2x−1)2 is equal to ______.

Correct answer: 5

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2025
cos⁡(sin⁡−135+sin⁡−1513+sin⁡−13365)\cos\left(\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}+\sin^{-1}\frac{33}{65}\right)cos(sin−153​+sin−1135​+sin−16533​) is equal to:
  1. (A)1
  2. (B)0
  3. (C)3365\frac{33}{65}6533​
  4. (D)3265\frac{32}{65}6532​

Correct answer: (B)

Step-by-step solution →
Q20·MathematicsSingle correctJEE Main 2025
Let [x][x][x] denote the greatest integer less than or equal to x. Then domain of f(x)=sec⁡−1(2[x]+1)f(x)=\sec^{-1}(2[x]+1)f(x)=sec−1(2[x]+1) is:
  1. (A)(−∞,−1]∪[0,∞)(-\infty,-1]\cup[0,\infty)(−∞,−1]∪[0,∞)
  2. (B)(−∞,∞)(-\infty,\infty)(−∞,∞)
  3. (C)(−∞,−1]∪[1,∞)(-\infty,-1]\cup[1,\infty)(−∞,−1]∪[1,∞)
  4. (D)(−∞,∞)−{0}(-\infty,\infty)-\{0\}(−∞,∞)−{0}

Correct answer: (B)

Step-by-step solution →
Q21·MathematicsIntegerJEE Main 2025
If for some α,β\alpha,\betaα,β; α≤β\alpha\le\betaα≤β, α+β=8\alpha+\beta=8α+β=8 and sec⁡2(tan⁡−1α)+cosec⁡2(cot⁡−1β)=36\sec^2(\tan^{-1}\alpha)+\operatorname{cosec}^2(\cot^{-1}\beta)=36sec2(tan−1α)+cosec2(cot−1β)=36, then α2+β\alpha^2+\betaα2+β is _______

Correct answer: 14

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2025
If α>β>γ>0\alpha>\beta>\gamma>0α>β>γ>0, then the expression cot⁡−1 ⁣{β+1+β2α−β}+cot⁡−1 ⁣{γ+1+γ2β−γ}+cot⁡−1 ⁣{α+1+α2γ−α}\cot^{-1}\!\left\{\beta+\dfrac{1+\beta^2}{\alpha-\beta}\right\}+\cot^{-1}\!\left\{\gamma+\dfrac{1+\gamma^2}{\beta-\gamma}\right\}+\cot^{-1}\!\left\{\alpha+\dfrac{1+\alpha^2}{\gamma-\alpha}\right\}cot−1{β+α−β1+β2​}+cot−1{γ+β−γ1+γ2​}+cot−1{α+γ−α1+α2​} is equal to:
  1. (A)π2−(α+β+γ)\dfrac{\pi}{2}-(\alpha+\beta+\gamma)2π​−(α+β+γ)
  2. (B)3π3\pi3π
  3. (C)000
  4. (D)π\piπ

Correct answer: (D)

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2025
If π2≤x≤3π4\dfrac{\pi}{2}\le x\le\dfrac{3\pi}{4}2π​≤x≤43π​, then cos⁡−1(1213cos⁡x+513sin⁡x)\cos^{-1}\left(\dfrac{12}{13}\cos x+\dfrac{5}{13}\sin x\right)cos−1(1312​cosx+135​sinx) is equal to
  1. (A)x−tan⁡−143x-\tan^{-1}\dfrac{4}{3}x−tan−134​
  2. (B)x−tan⁡−1512x-\tan^{-1}\dfrac{5}{12}x−tan−1125​
  3. (C)x+tan⁡−145x+\tan^{-1}\dfrac{4}{5}x+tan−154​
  4. (D)x+tan⁡−1512x+\tan^{-1}\dfrac{5}{12}x+tan−1125​

Correct answer: (B)

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2025
Using principal values, the sum of the maximum and minimum values of 16[(sec⁡−1x)2+(cosec⁡−1x)2]16\big[(\sec^{-1}x)^2+(\operatorname{cosec}^{-1}x)^2\big]16[(sec−1x)2+(cosec−1x)2] is:
  1. (A)24π224\pi^224π2
  2. (B)18π218\pi^218π2
  3. (C)31π231\pi^231π2
  4. (D)22π222\pi^222π2

Correct answer: (D)

Step-by-step solution →
Q25·MathematicsSingle correctJEE Advanced 2024
Considering only the principal values of the inverse trigonometric functions, the value of tan⁡(sin⁡−1(35)−2cos⁡−1(25))\tan\left(\sin^{-1}\left(\frac{3}{5}\right) - 2\cos^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)tan(sin−1(53​)−2cos−1(5​2​)) is
  1. (A)724\frac{7}{24}247​
  2. (B)−724\frac{-7}{24}24−7​
  3. (C)−524\frac{-5}{24}24−5​
  4. (D)524\frac{5}{24}245​

Correct answer: (B)

Step-by-step solution →
Q26·MathematicsNumericalJEE Main 2024
Let the inverse trigonometric functions take principal values. The number of real solutions of the equation 2sin⁡−1x+3cos⁡−1x=2π52\sin^{-1}x+3\cos^{-1}x=\dfrac{2\pi}{5}2sin−1x+3cos−1x=52π​, is _______.

Correct answer: 0

Step-by-step solution →
Q27·MathematicsNumericalJEE Main 2024
For n∈Nn\in\mathbb{N}n∈N, if cot⁡−13+cot⁡−14+cot⁡−15+cot⁡−1n=π4\cot^{-1}3+\cot^{-1}4+\cot^{-1}5+\cot^{-1}n=\dfrac{\pi}{4}cot−13+cot−14+cot−15+cot−1n=4π​, then nnn is equal to _______.

Correct answer: 47

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2024
Given the inverse trigonometric function assumes principal values only. Let xxx, yyy be any two real numbers in [−1,1][-1,1][−1,1] such that cos⁡−1x−sin⁡−1y=α\cos^{-1}x-\sin^{-1}y=\alphacos−1x−sin−1y=α, −π2≤α≤π-\dfrac{\pi}{2}\le\alpha\le\pi−2π​≤α≤π. Then, the minimum value of x2+y2+2xysin⁡αx^2+y^2+2xy\sin\alphax2+y2+2xysinα is
  1. (A)−1-1−1
  2. (B)000
  3. (C)−12-\tfrac{1}{2}−21​
  4. (D)12\tfrac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2024
If a=sin⁡−1(sin⁡(5))a=\sin^{-1}(\sin(5))a=sin−1(sin(5)) and b=cos⁡−1(cos⁡(5))b=\cos^{-1}(\cos(5))b=cos−1(cos(5)), then a2+b2a^2+b^2a2+b2 is equal to
  1. (A)4π2+254\pi^2+254π2+25
  2. (B)8π2−40π+508\pi^2-40\pi+508π2−40π+50
  3. (C)4π2−20π+504\pi^2-20\pi+504π2−20π+50
  4. (D)25

Correct answer: (B)

Step-by-step solution →
Q30·MathematicsSingle correctJEE Main 2024
For α,β,γ≠0\alpha,\beta,\gamma\ne 0α,β,γ=0. If sin⁡−1α+sin⁡−1β+sin⁡−1γ=π\sin^{-1}\alpha+\sin^{-1}\beta+\sin^{-1}\gamma=\pisin−1α+sin−1β+sin−1γ=π and (α+β+γ)(α−γ+β)=3αβ(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3\alpha\beta(α+β+γ)(α−γ+β)=3αβ, then γ\gammaγ equal to
  1. (A)32\dfrac{\sqrt{3}}{2}23​​
  2. (B)12\dfrac{1}{\sqrt{2}}2​1​
  3. (C)3−122\dfrac{\sqrt{3}-1}{2\sqrt{2}}22​3​−1​
  4. (D)3\sqrt{3}3​

Correct answer: (A)

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2024
Let x=mnx=\dfrac{m}{n}x=nm​ (mmm, nnn are co-prime natural numbers) be a solution of the equation cos⁡(2sin⁡−1x)=19\cos\left(2\sin^{-1}x\right)=\dfrac{1}{9}cos(2sin−1x)=91​ and let α,β (α>β)\alpha,\beta\,(\alpha>\beta)α,β(α>β) be the roots of the equation mx2−nx−m+n=0mx^2-nx-m+n=0mx2−nx−m+n=0. Then the point (α,β)(\alpha,\beta)(α,β) lies on the line:
  1. (A)3x+2y=23x+2y=23x+2y=2
  2. (B)5x−8y=−95x-8y=-95x−8y=−9
  3. (C)3x−2y=−23x-2y=-23x−2y=−2
  4. (D)5x+8y=95x+8y=95x+8y=9

Correct answer: (D)

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2024
Considering only the principal values of inverse trigonometric functions, the number of positive real values of xxx satisfying tan⁡−1(x)+tan⁡−1(2x)=π4\tan^{-1}(x)+\tan^{-1}(2x)=\dfrac{\pi}{4}tan−1(x)+tan−1(2x)=4π​ is :
  1. (A)More than 2
  2. (B)1
  3. (C)2
  4. (D)0

Correct answer: (B)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Advanced 2023
For any y∈Ry \in \mathbb{R}y∈R, let cot⁡−1(y)∈(0,π)\cot^{-1}(y) \in (0, \pi)cot−1(y)∈(0,π) and tan⁡−1(y)∈(−π2,π2)\tan^{-1}(y) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)tan−1(y)∈(−2π​,2π​). Then the sum of all the solutions of the equation tan⁡−1(6y9−y2)+cot⁡−1(9−y26y)=2π3\tan^{-1}\left(\frac{6y}{9-y^{2}}\right) + \cot^{-1}\left(\frac{9-y^{2}}{6y}\right) = \frac{2\pi}{3}tan−1(9−y26y​)+cot−1(6y9−y2​)=32π​ for 0<∣y∣<30 < |y| < 30<∣y∣<3, is equal to
  1. (A)23−32\sqrt{3} - 323​−3
  2. (B)3−233 - 2\sqrt{3}3−23​
  3. (C)43−64\sqrt{3} - 643​−6
  4. (D)6−436 - 4\sqrt{3}6−43​

Correct answer: (C)

Step-by-step solution →
Q34·MathematicsNumericalJEE Advanced 2023
Let tan⁡−1(x)∈(−π2,π2)\tan^{-1}(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)tan−1(x)∈(−2π​,2π​), for x∈Rx \in Rx∈R. Then the number of real solutions of the equation 1+cos⁡(2x)=2tan⁡−1(tan⁡x)\sqrt{1 + \cos(2x)} = \sqrt{2}\tan^{-1}(\tan x)1+cos(2x)​=2​tan−1(tanx) in the set (−3π2,−π2)∪(−π2,π2)∪(π2,3π2)\left(-\frac{3\pi}{2}, -\frac{\pi}{2}\right) \cup \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{2}\right)(−23π​,−2π​)∪(−2π​,2π​)∪(2π​,23π​) is equal to

Correct answer: 3

Step-by-step solution →
Q35·MathematicsNumericalJEE Main 2023
For x∈(−1,1]x \in (-1, 1]x∈(−1,1], the number of solutions of the equation sin⁡−1x=2tan⁡−1x\sin^{-1}x = 2\tan^{-1}xsin−1x=2tan−1x is equal to

Correct answer: 2

Step-by-step solution →
Q36·MathematicsNumericalJEE Main 2023
Let S={x∈R:sin⁡−1(x+1x2+2x+2)−sin⁡−1(xx2+1)=π4}S=\left\{x\in\mathbb{R}:\sin^{-1}\left(\dfrac{x+1}{\sqrt{x^{2}+2x+2}}\right)-\sin^{-1}\left(\dfrac{x}{\sqrt{x^{2}+1}}\right)=\dfrac{\pi}{4}\right\}S={x∈R:sin−1(x2+2x+2​x+1​)−sin−1(x2+1​x​)=4π​}. Then ∑x∈S(sin⁡ ⁣((x2+x+5)π2)−cos⁡ ⁣((x2+x+5)π))\displaystyle\sum_{x\in S}\left(\sin\!\left((x^{2}+x+5)\dfrac{\pi}{2}\right)-\cos\!\left((x^{2}+x+5)\pi\right)\right)x∈S∑​(sin((x2+x+5)2π​)−cos((x2+x+5)π)) is equal to _____.

Correct answer: 4

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2023
The range of f(x)=4sin⁡−1(x2x2+1)f(x) = 4\sin^{-1}\left(\dfrac{x^2}{x^2+1}\right)f(x)=4sin−1(x2+1x2​) is
  1. (A)[0,π][0, \pi][0,π]
  2. (B)[0,2π)[0, 2\pi)[0,2π)
  3. (C)[0,π)[0, \pi)[0,π)
  4. (D)[0,2π][0, 2\pi][0,2π]

Correct answer: (B)

Step-by-step solution →
Q38·MathematicsNumericalJEE Main 2023
If the domain of the function f(x)=sec⁡−1(2x5x+3)f(x)=\sec^{-1}\left(\frac{2x}{5x+3}\right)f(x)=sec−1(5x+32x​) is [α,β)∪(γ,δ][\alpha,\beta)\cup(\gamma,\delta][α,β)∪(γ,δ], then ∣3α+10(β+γ)+21δ∣\left|3\alpha+10(\beta+\gamma)+21\delta\right|∣3α+10(β+γ)+21δ∣ is equal to _______ .

Correct answer: 24

Step-by-step solution →
Q39·MathematicsNumericalJEE Main 2023
If domain of the function log⁡e(6x2+5x+12x−1)+cos⁡−1(2x2−3x+43x−5)\log_e\left(\dfrac{6x^{2}+5x+1}{2x-1}\right)+\cos^{-1}\left(\dfrac{2x^{2}-3x+4}{3x-5}\right)loge​(2x−16x2+5x+1​)+cos−1(3x−52x2−3x+4​) is (α,β)∪(γ,δ](\alpha,\beta)\cup(\gamma,\delta](α,β)∪(γ,δ], then 18(α2+β2+γ2+δ2)18(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2})18(α2+β2+γ2+δ2) is equal to

Correct answer: 20

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2023
Let S={x∈R:0<x<1 and 2tan⁡−1(1−x1+x)=cos⁡−1(1−x21+x2)}S=\left\{x\in\mathbb{R}:0<x<1\text{ and }2\tan^{-1}\left(\dfrac{1-x}{1+x}\right)=\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)\right\}S={x∈R:0<x<1 and 2tan−1(1+x1−x​)=cos−1(1+x21−x2​)}. If n(S)n(S)n(S) denotes the number of elements in S then:
  1. (A)n(S)=2n(S)=2n(S)=2 and only one element in S is less than 12\dfrac1221​.
  2. (B)n(S)=1n(S)=1n(S)=1 and the element in S is more than 12\dfrac1221​.
  3. (C)n(S)=0n(S)=0n(S)=0
  4. (D)n(S)=1n(S)=1n(S)=1 and the element in S is less than 12\dfrac1221​.

Correct answer: (A)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2023
Let SSS be the set of all solutions of the equation cos⁡−1(2x)−2cos⁡−1 ⁣(1−x2)=π\cos^{-1}(2x)-2\cos^{-1}\!\left(\sqrt{1-x^2}\right)=\picos−1(2x)−2cos−1(1−x2​)=π, x∈[−12,12]x\in\left[-\tfrac{1}{2},\tfrac{1}{2}\right]x∈[−21​,21​]. Then ∑x∈S2sin⁡−1 ⁣(x2−1)\sum_{x\in S}2\sin^{-1}\!\left(x^2-1\right)∑x∈S​2sin−1(x2−1) is equal to
  1. (A)π−2sin⁡−1 ⁣(34)\pi-2\sin^{-1}\!\left(\frac{\sqrt{3}}{4}\right)π−2sin−1(43​​)
  2. (B)π−sin⁡−1 ⁣(34)\pi-\sin^{-1}\!\left(\frac{\sqrt{3}}{4}\right)π−sin−1(43​​)
  3. (C)−2π3\frac{-2\pi}{3}3−2π​
  4. (D)000

Correct answer: (A)

Step-by-step solution →
Q42·MathematicsSingle correctJEE Main 2023
Let (a,b)⊂(0,2π)(a,b)\subset(0,2\pi)(a,b)⊂(0,2π) be the largest interval for which sin⁡−1(sin⁡θ)−cos⁡−1(sin⁡θ)>0, θ∈(0,2π)\sin^{-1}(\sin\theta)-\cos^{-1}(\sin\theta)>0,\ \theta\in(0,2\pi)sin−1(sinθ)−cos−1(sinθ)>0, θ∈(0,2π), holds. If αx2+βx+sin⁡−1(x2−6x+10)+cos⁡−1(x2−6x+10)=0\alpha x^2+\beta x+\sin^{-1}(x^2-6x+10)+\cos^{-1}(x^2-6x+10)=0αx2+βx+sin−1(x2−6x+10)+cos−1(x2−6x+10)=0 and α−β=b−a\alpha-\beta=b-aα−β=b−a, then α\alphaα is equal to:
  1. (A)π16\dfrac{\pi}{16}16π​
  2. (B)π48\dfrac{\pi}{48}48π​
  3. (C)π12\dfrac{\pi}{12}12π​
  4. (D)π8\dfrac{\pi}{8}8π​

Correct answer: (C)

Step-by-step solution →
Q43·MathematicsSingle correctJEE Main 2023
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 < 13sin−117α​+cos−154​−tan−13677​=0, 0<α<13, then sin⁡−1(sin⁡α)+cos⁡−1(cos⁡α)\sin^{-1}(\sin \alpha) + \cos^{-1}(\cos \alpha)sin−1(sinα)+cos−1(cosα) is equal to
  1. (A)161616
  2. (B)000
  3. (C)π\piπ
  4. (D)16−5π16 - 5\pi16−5π

Correct answer: (C)

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2023
Let y=f(x)y=f(x)y=f(x) represent a parabola with focus (−12,0)\left(-\frac{1}{2},0\right)(−21​,0) and directrix y=−12y=-\frac{1}{2}y=−21​. Then S={x∈R:tan⁡−1(f(x))+sin⁡−1(f(x)+1)=π2}S=\left\{x\in\mathbb{R}:\tan^{-1}(\sqrt{f(x)})+\sin^{-1}(\sqrt{f(x)+1})=\frac{\pi}{2}\right\}S={x∈R:tan−1(f(x)​)+sin−1(f(x)+1​)=2π​}:
  1. (A)contains exactly two elements
  2. (B)contains exactly one element
  3. (C)is an empty set
  4. (D)is an infinite set

Correct answer: (A)

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2023
Let a1=1,a2,a3,a4,…a_1=1, a_2, a_3, a_4, \ldotsa1​=1,a2​,a3​,a4​,… be consecutive natural numbers. Then tan⁡−1(11+a1a2)+tan⁡−1(11+a2a3)+⋯+tan⁡−1(11+a2021a2022)\tan^{-1}\left(\dfrac{1}{1 + a_1 a_2}\right) + \tan^{-1}\left(\dfrac{1}{1 + a_2 a_3}\right) + \cdots + \tan^{-1}\left(\dfrac{1}{1 + a_{2021} a_{2022}}\right)tan−1(1+a1​a2​1​)+tan−1(1+a2​a3​1​)+⋯+tan−1(1+a2021​a2022​1​) is equal to:
  1. (A)cot⁡−1(2022)−π4\cot^{-1}(2022) - \dfrac{\pi}{4}cot−1(2022)−4π​
  2. (B)π4−cot⁡−1(2022)\dfrac{\pi}{4} - \cot^{-1}(2022)4π​−cot−1(2022)
  3. (C)tan⁡−1(2022)−π4\tan^{-1}(2022) - \dfrac{\pi}{4}tan−1(2022)−4π​
  4. (D)π4−tan⁡−1(2022)\dfrac{\pi}{4} - \tan^{-1}(2022)4π​−tan−1(2022)

Correct answer: (C)

Step-by-step solution →
Q46·MathematicsNumericalJEE Main 2023
If the sum of all the solutions of tan⁡−1(2x1−x2)+cot⁡−1(1−x22x)=π3\tan^{-1}\left(\dfrac{2x}{1-x^2}\right)+\cot^{-1}\left(\dfrac{1-x^2}{2x}\right)=\dfrac{\pi}{3}tan−1(1−x22x​)+cot−1(2x1−x2​)=3π​, −1<x<1-1<x<1−1<x<1, x≠0x\ne 0x=0, is α−43\alpha-\dfrac{4}{\sqrt{3}}α−3​4​, then α\alphaα is equal to _______.

Correct answer: 2

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2023
tan⁡−1 ⁣(1+33+3)+sec⁡−1 ⁣(8+436+33)\tan^{-1}\!\left(\dfrac{1+\sqrt{3}}{3+\sqrt{3}}\right)+\sec^{-1}\!\left(\sqrt{\dfrac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right)tan−1(3+3​1+3​​)+sec−1(6+33​8+43​​​) is equal to:
  1. (A)π3\tfrac{\pi}{3}3π​
  2. (B)π4\tfrac{\pi}{4}4π​
  3. (C)π6\tfrac{\pi}{6}6π​
  4. (D)π2\tfrac{\pi}{2}2π​

Correct answer: (A)

Step-by-step solution →
Q48·MathematicsSingle correctJEE Main 2022
Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos⁡−1(x)−2sin⁡−1(x)=cos⁡−1(2x)\cos^{-1}(x) - 2\sin^{-1}(x) = \cos^{-1}(2x)cos−1(x)−2sin−1(x)=cos−1(2x) is equal to :
  1. (A)0
  2. (B)1
  3. (C)12\frac{1}{2}21​
  4. (D)−12-\frac{1}{2}−21​

Correct answer: (A)

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2022
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}-4x+2}{x^{2}+3}\right)f(x)=cos−1(x2+3x2−4x+2​) is :
  1. (A)(−∞,14]\left(-\infty, \frac{1}{4}\right](−∞,41​]
  2. (B)[−14,∞)\left[-\frac{1}{4}, \infty\right)[−41​,∞)
  3. (C)(−13,∞)\left(-\frac{1}{3}, \infty\right)(−31​,∞)
  4. (D)(−∞,13]\left(-\infty, \frac{1}{3}\right](−∞,31​]

Correct answer: (B)

Step-by-step solution →
Q50·MathematicsNumericalJEE Main 2022
For k∈Rk\in\mathbb{R}k∈R, let the solutions of the equation cos⁡(sin⁡−1(xcot⁡(tan⁡−1(cos⁡(sin⁡−1x)))))=k,  0<∣x∣<12\cos\left(\sin^{-1}\left(x\cot\left(\tan^{-1}\left(\cos\left(\sin^{-1}x\right)\right)\right)\right)\right)=k,\;0<\left|x\right|<\frac{1}{\sqrt{2}}cos(sin−1(xcot(tan−1(cos(sin−1x)))))=k,0<∣x∣<2​1​ be α\alphaα and β\betaβ, where the inverse trigonometric functions take only principal values. If the solutions of the equation x2−bx−5=0x^{2}-bx-5=0x2−bx−5=0 are 1α2+1β2\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}α21​+β21​ and αβ\frac{\alpha}{\beta}βα​, then bk2\frac{b}{k^{2}}k2b​ is equal to ______.

Correct answer: 12

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2022
If 0<x<120<x<\frac{1}{\sqrt{2}}0<x<2​1​ and sin⁡−1xα=cos⁡−1xβ\frac{\sin^{-1}x}{\alpha}=\frac{\cos^{-1}x}{\beta}αsin−1x​=βcos−1x​, then a value of sin⁡(2παα+β)\sin\left(\frac{2\pi\alpha}{\alpha+\beta}\right)sin(α+β2πα​) is
  1. (A)4(1−x2)(1−2x2)4\sqrt{\left(1-x^{2}\right)}\left(1-2x^{2}\right)4(1−x2)​(1−2x2)
  2. (B)4x(1−x2)(1−2x2)4x\sqrt{\left(1-x^{2}\right)}\left(1-2x^{2}\right)4x(1−x2)​(1−2x2)
  3. (C)2x(1−x2)(1−4x2)2x\sqrt{\left(1-x^{2}\right)}\left(1-4x^{2}\right)2x(1−x2)​(1−4x2)
  4. (D)4(1−x2)(1−4x2)4\sqrt{\left(1-x^{2}\right)}\left(1-4x^{2}\right)4(1−x2)​(1−4x2)

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2022
tan⁡(2tan⁡−115+sec⁡−152+2tan⁡−118)\tan\left( 2\tan^{-1}\frac{1}{5} + \sec^{-1}\frac{\sqrt{5}}{2} + 2\tan^{-1}\frac{1}{8} \right)tan(2tan−151​+sec−125​​+2tan−181​) is equal to:
  1. (A)1
  2. (B)2
  3. (C)14\frac{1}{4}41​
  4. (D)54\frac{5}{4}45​

Correct answer: (B)

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2022
The domain of the function cos⁡−1(2sin⁡−1(14x2−1)π)\cos^{-1}\left(\frac{2\sin^{-1}\left(\frac{1}{4x^{2} - 1}\right)}{\pi}\right)cos−1(π2sin−1(4x2−11​)​) is :
  1. (A)R−{−12,12}R - \left\{-\frac{1}{2}, \frac{1}{2}\right\}R−{−21​,21​}
  2. (B)(−∞,−1]∪[1,∞)∪{0}(-\infty, -1] \cup [1, \infty) \cup \{0\}(−∞,−1]∪[1,∞)∪{0}
  3. (C)(−∞,−12)∪(12,∞)∪{0}\left(-\infty, \frac{-1}{2}\right) \cup \left(\frac{1}{2}, \infty\right) \cup \{0\}(−∞,2−1​)∪(21​,∞)∪{0}
  4. (D)(−∞,−12]∪[12,∞)∪{0}\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup \left[\frac{1}{\sqrt{2}}, \infty\right) \cup \{0\}(−∞,2​−1​]∪[2​1​,∞)∪{0}

Correct answer: (D)

Step-by-step solution →
Q54·MathematicsNumericalJEE Main 2022
50tan⁡(3tan⁡−1(12)+2cos⁡−1(15))+42tan⁡(12tan⁡−1(22))50\tan\left(3\tan^{-1}\left(\frac{1}{2}\right) + 2\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)\right) + 4\sqrt{2}\tan\left(\frac{1}{2}\tan^{-1}(2\sqrt{2})\right)50tan(3tan−1(21​)+2cos−1(5​1​))+42​tan(21​tan−1(22​)) is equal to _____.

Correct answer: 29

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2022
The value of lim⁡n→∞6tan⁡{∑r=1ntan⁡−1(1r2+3r+3)}\lim\limits_{n\to\infty} 6\tan\left\{\sum\limits_{r=1}^{n} \tan^{-1}\left(\frac{1}{r^2 + 3r + 3}\right)\right\}n→∞lim​6tan{r=1∑n​tan−1(r2+3r+31​)} is equal to
  1. (A)111
  2. (B)222
  3. (C)333
  4. (D)666

Correct answer: (C)

Step-by-step solution →
Q56·MathematicsSingle correctJEE Main 2022
The value of cot⁡(∑n=150tan⁡−1(11+n+n2))\cot\left(\sum_{n=1}^{50}\tan^{-1}\left(\dfrac{1}{1+n+n^{2}}\right)\right)cot(∑n=150​tan−1(1+n+n21​)) is
  1. (A)2625\dfrac{26}{25}2526​
  2. (B)2526\dfrac{25}{26}2625​
  3. (C)5051\dfrac{50}{51}5150​
  4. (D)5251\dfrac{52}{51}5152​

Correct answer: (A)

Step-by-step solution →
Q57·MathematicsSingle correctJEE Main 2022
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)sin−1(sin32π​)+cos−1(cos67π​)+tan−1(tan43π​) is equal to :
  1. (A)11π12\frac{11\pi}{12}1211π​
  2. (B)17π12\frac{17\pi}{12}1217π​
  3. (C)31π12\frac{31\pi}{12}1231π​
  4. (D)−3π4-\frac{3\pi}{4}−43π​

Correct answer: (A)

Step-by-step solution →
Q58·MathematicsSingle correctJEE Main 2022
Let f(x)=2cos⁡−1x+4cot⁡−1x−3x2−2x+10f(x) = 2\cos^{-1}x + 4\cot^{-1}x - 3x^{2} - 2x + 10f(x)=2cos−1x+4cot−1x−3x2−2x+10, x∈[−1,1]x \in [-1, 1]x∈[−1,1]. If [a,b][a, b][a,b] is the range of the function then 4a−b4a - b4a−b is equal to:
  1. (A)111111
  2. (B)11−π11 - \pi11−π
  3. (C)11+π11 + \pi11+π
  4. (D)15−π15 - \pi15−π

Correct answer: (B)

Step-by-step solution →
Q59·MathematicsSingle correctJEE Main 2022
If the inverse trigonometric functions take principal values, then cos⁡−1(310cos⁡(tan⁡−1(43))+25sin⁡(tan⁡−1(43)))\cos^{-1}\left(\frac{3}{10}\cos\left(\tan^{-1}\left(\frac{4}{3}\right)\right)+\frac{2}{5}\sin\left(\tan^{-1}\left(\frac{4}{3}\right)\right)\right)cos−1(103​cos(tan−1(34​))+52​sin(tan−1(34​))) is equal to :
  1. (A)0
  2. (B)π4\frac{\pi}{4}4π​
  3. (C)π3\frac{\pi}{3}3π​
  4. (D)π6\frac{\pi}{6}6π​

Correct answer: (C)

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2022
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)tan−1(sin(4π​)cos(415π​)−1​) is equal to
  1. (A)−π4-\frac{\pi}{4}−4π​
  2. (B)−π8-\frac{\pi}{8}−8π​
  3. (C)−5π12-\frac{5\pi}{12}−125π​
  4. (D)−4π9-\frac{4\pi}{9}−94π​

Correct answer: (B)

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2022
The set of all values of k for which (tan⁡−1x)3+(cot⁡−1x)3=kπ3(\tan^{-1}x)^{3}+(\cot^{-1}x)^{3}=k\pi^{3}(tan−1x)3+(cot−1x)3=kπ3, x∈Rx\in Rx∈R, is the interval :
  1. (A)[132,78)\left[\dfrac{1}{32},\dfrac{7}{8}\right)[321​,87​)
  2. (B)(124,1316)\left(\dfrac{1}{24},\dfrac{13}{16}\right)(241​,1613​)
  3. (C)[148,1316]\left[\dfrac{1}{48},\dfrac{13}{16}\right][481​,1613​]
  4. (D)[132,98)\left[\dfrac{1}{32},\dfrac{9}{8}\right)[321​,89​)

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsSingle correctJEE Main 2022
Let x∗y=x2+y3x*y = x^{2} + y^{3}x∗y=x2+y3 and (x∗1)∗1=x∗(1∗1)(x*1)*1 = x*(1*1)(x∗1)∗1=x∗(1∗1). Then a value of 2sin⁡−1(x4+x2−2x4+x2+2)2\sin^{-1}\left(\dfrac{x^{4}+x^{2}-2}{x^{4}+x^{2}+2}\right)2sin−1(x4+x2+2x4+x2−2​) is
  1. (A)π4\dfrac{\pi}{4}4π​
  2. (B)π3\dfrac{\pi}{3}3π​
  3. (C)π2\dfrac{\pi}{2}2π​
  4. (D)π6\dfrac{\pi}{6}6π​

Correct answer: (B)

Step-by-step solution →
Q63·MathematicsMultiple correctJEE Advanced 2021
For any positive integer n, let Sn:(0,∞)→RS_n : (0, \infty) \to \mathbb{R}Sn​:(0,∞)→R be defined by Sn(x)=∑k=1ncot⁡−1(1+k(k+1)x2x)S_n(x) = \sum_{k=1}^{n} \cot^{-1}\left(\frac{1 + k(k+1)x^2}{x}\right)Sn​(x)=∑k=1n​cot−1(x1+k(k+1)x2​), where for any x∈Rx \in \mathbb{R}x∈R, cot⁡−1x∈(0,π)\cot^{-1}x \in (0, \pi)cot−1x∈(0,π) and tan⁡−1(x)∈(−π2,π2)\tan^{-1}(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)tan−1(x)∈(−2π​,2π​). Then which of the following statements is (are) TRUE ?
  1. (A)S10(x)=π2−tan⁡−1(1+11x210x)S_{10}(x) = \frac{\pi}{2} - \tan^{-1}\left(\frac{1 + 11x^2}{10x}\right)S10​(x)=2π​−tan−1(10x1+11x2​), for all x>0x > 0x>0
  2. (B)lim⁡n→∞cot⁡(Sn(x))=x\lim_{n \to \infty} \cot\left(S_n(x)\right) = xlimn→∞​cot(Sn​(x))=x, for all x>0x > 0x>0
  3. (C)The equation S3(x)=π4S_3(x) = \frac{\pi}{4}S3​(x)=4π​ has a root in (0,∞)(0, \infty)(0,∞)
  4. (D)tan⁡(Sn(x))≤12\tan\left(S_n(x)\right) \le \frac{1}{2}tan(Sn​(x))≤21​, for all n≥1n \ge 1n≥1 and x>0x > 0x>0

Correct answer: (A), (B)

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2021
cos⁡−1(cos⁡(−5))+sin⁡−1(sin⁡(6))−tan⁡−1(tan⁡(12))\cos^{-1}(\cos(-5)) + \sin^{-1}(\sin(6)) - \tan^{-1}(\tan(12))cos−1(cos(−5))+sin−1(sin(6))−tan−1(tan(12)) is equal to : (The inverse trigonometric functions take the principal values)
  1. (A)3π − 11
  2. (B)4 π − 9
  3. (C)4 π − 11
  4. (D)3π + 1

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2021
The domain of the function f(x)=sin⁡−1(3x2+x−1(x−1)2)+cos⁡−1(x−1x+1)f(x) = \sin^{-1}\left(\frac{3x^2 + x - 1}{(x - 1)^2}\right) + \cos^{-1}\left(\frac{x - 1}{x + 1}\right)f(x)=sin−1((x−1)23x2+x−1​)+cos−1(x+1x−1​) is :
  1. (A)[0,14]\left[0, \frac{1}{4}\right][0,41​]
  2. (B)[−2,0]∪[14,12][-2, 0] \cup \left[\frac{1}{4}, \frac{1}{2}\right][−2,0]∪[41​,21​]
  3. (C)[14,12]∪{0}\left[\frac{1}{4}, \frac{1}{2}\right] \cup \{0\}[41​,21​]∪{0}
  4. (D)[0,12]\left[0, \frac{1}{2}\right][0,21​]

Correct answer: (C)

Step-by-step solution →
Q66·MathematicsSingle correctJEE Main 2021
If y(x)=cot⁡−1(1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x)y(x) = \cot^{-1}\left(\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\right)y(x)=cot−1(1+sinx​−1−sinx​1+sinx​+1−sinx​​), x ∈ (π2,π)\left(\frac{\pi}{2}, \pi\right)(2π​,π), then dydx\frac{dy}{dx}dxdy​ at x=5π6x = \frac{5\pi}{6}x=65π​ is:
  1. (A)−12-\frac{1}{2}−21​
  2. (B)−1
  3. (C)12\frac{1}{2}21​
  4. (D)0

Correct answer: (A)

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2021
If (sin⁡−1x)2−(cos⁡−1x)2=a(\sin^{-1} x)^2 - (\cos^{-1} x)^2 = a(sin−1x)2−(cos−1x)2=a; 0<x<10 < x < 10<x<1, a≠0a \neq 0a=0, then the value of 2x2−12x^2 - 12x2−1 is :
  1. (A)cos⁡(4aπ)\cos\left(\frac{4a}{\pi}\right)cos(π4a​)
  2. (B)sin⁡(2aπ)\sin\left(\frac{2a}{\pi}\right)sin(π2a​)
  3. (C)cos⁡(2aπ)\cos\left(\frac{2a}{\pi}\right)cos(π2a​)
  4. (D)sin⁡(4aπ)\sin\left(\frac{4a}{\pi}\right)sin(π4a​)

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correctJEE Main 2021
The domain of the function cosec−1(1+xx)\mathrm{cosec}^{-1}\left(\frac{1 + x}{x}\right)cosec−1(x1+x​) is :
  1. (A)(−1,−12]∪(0,∞)\left(-1, -\frac{1}{2}\right] \cup (0, \infty)(−1,−21​]∪(0,∞)
  2. (B)[−12,0)∪[1,∞)\left[-\frac{1}{2}, 0\right) \cup [1, \infty)[−21​,0)∪[1,∞)
  3. (C)(−12,∞)−{0}\left(-\frac{1}{2}, \infty\right) - \{0\}(−21​,∞)−{0}
  4. (D)[−12,∞)−{0}\left[-\frac{1}{2}, \infty\right) - \{0\}[−21​,∞)−{0}

Correct answer: (D)

Step-by-step solution →
Q69·MathematicsSingle correctJEE Main 2021
Let f(x)=cos⁡(2tan⁡−1sin⁡(cot⁡−11−xx))f(x) = \cos\left(2\tan^{-1}\sin\left(\cot^{-1}\sqrt{\frac{1-x}{x}}\right)\right)f(x)=cos(2tan−1sin(cot−1x1−x​​)), 0<x<10 < x < 10<x<1. Then :
  1. (A)(1−x)2f′(x)−2(f(x))2=0(1 - x)^2 f'(x) - 2(f(x))^2 = 0(1−x)2f′(x)−2(f(x))2=0
  2. (B)(1+x)2f′(x)+2(f(x))2=0(1 + x)^2 f'(x) + 2(f(x))^2 = 0(1+x)2f′(x)+2(f(x))2=0
  3. (C)(1−x)2f′(x)+2(f(x))2=0(1 - x)^2 f'(x) + 2(f(x))^2 = 0(1−x)2f′(x)+2(f(x))2=0
  4. (D)(1+x)2f′(x)−2(f(x))2=0(1 + x)^2 f'(x) - 2(f(x))^2 = 0(1+x)2f′(x)−2(f(x))2=0

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2021
If ∑r=150tan⁡−112r2=p\sum_{r=1}^{50} \tan^{-1} \frac{1}{2r^{2}} = p∑r=150​tan−12r21​=p, then the value of tan⁡p\tan ptanp is :
  1. (A)101102\frac{101}{102}102101​
  2. (B)5051\frac{50}{51}5150​
  3. (C)100100100
  4. (D)5150\frac{51}{50}5051​

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2021
If the domain of the function f(x)=cos⁡−1x2−x+1sin⁡−1(2x−12)f(x) = \dfrac{\cos^{-1}\sqrt{x^2 - x + 1}}{\sqrt{\sin^{-1}\left(\dfrac{2x - 1}{2}\right)}}f(x)=sin−1(22x−1​)​cos−1x2−x+1​​ is the interval (α,β]\left(\alpha, \beta\right](α,β], then α+β\alpha + \betaα+β is equal to :
  1. (A)111
  2. (B)12\dfrac{1}{2}21​
  3. (C)222
  4. (D)32\dfrac{3}{2}23​

Correct answer: (D)

Step-by-step solution →
Q72·MathematicsSingle correctJEE Main 2021
The number of real roots of the equation tan⁡−1x(x+1)+sin⁡−1x2+x+1=π4\tan^{-1}\sqrt{x(x+1)}+\sin^{-1}\sqrt{x^2+x+1} = \frac{\pi}{4}tan−1x(x+1)​+sin−1x2+x+1​=4π​ is :
  1. (A)4
  2. (B)1
  3. (C)0
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q73·MathematicsSingle correctJEE Main 2021
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)tan(2tan−1(53​)+sin−1(135​)) is equal to :
  1. (A)15163\frac{151}{63}63151​
  2. (B)−29176\frac{-291}{76}76−291​
  3. (C)22021\frac{220}{21}21220​
  4. (D)−18169\frac{-181}{69}69−181​

Correct answer: (C)

Step-by-step solution →
Q74·MathematicsSingle correctJEE Main 2021
The sum of possible values of x for tan⁡−1(x+1)+cot⁡−1(1x−1)=tan⁡−1(831)\tan^{-1}(x + 1) + \cot^{-1}\left(\frac{1}{x - 1}\right) = \tan^{-1}\left(\frac{8}{31}\right)tan−1(x+1)+cot−1(x−11​)=tan−1(318​) is :
  1. (A)−324-\frac{32}{4}−432​
  2. (B)−314-\frac{31}{4}−431​
  3. (C)−304-\frac{30}{4}−430​
  4. (D)−334-\frac{33}{4}−433​

Correct answer: (A)

Step-by-step solution →
Q75·MathematicsSingle correctJEE Main 2021
The number of solutions of the equation sin⁡−1[x2+13]+cos⁡−1[x2−23]=x2\sin^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos^{-1}\left[x^{2}-\frac{2}{3}\right] = x^{2}sin−1[x2+31​]+cos−1[x2−32​]=x2, for x∈[−1,1]x \in [-1, 1]x∈[−1,1], and [x][x][x] denotes the greatest integer less than or equal to xxx, is :
  1. (A)222
  2. (B)000
  3. (C)444
  4. (D)Infinite

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correctJEE Main 2021
If cot⁡−1(α)=cot⁡−12+cot⁡−18+cot⁡−118+cot⁡−132+.....\cot^{-1}(\alpha) = \cot^{-1} 2 + \cot^{-1} 8 + \cot^{-1} 18 + \cot^{-1} 32 + .....cot−1(α)=cot−12+cot−18+cot−118+cot−132+..... upto 100 terms, then α\alphaα is :
  1. (A)1.01
  2. (B)1.00
  3. (C)1.02
  4. (D)1.03

Correct answer: (A)

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2021
Let Sk=∑r=1ktan⁡−1(6r22r+1+32r+1)S_{k} = \sum_{r=1}^{k} \tan^{-1}\left(\frac{6^{r}}{2^{2r+1} + 3^{2r+1}}\right)Sk​=∑r=1k​tan−1(22r+1+32r+16r​). Then lim⁡k→∞Sk\lim_{k \to \infty} S_{k}limk→∞​Sk​ is equal to :
  1. (A)tan⁡−1(32)\tan^{-1}\left(\frac{3}{2}\right)tan−1(23​)
  2. (B)π2\frac{\pi}{2}2π​
  3. (C)cot⁡−1(32)\cot^{-1}\left(\frac{3}{2}\right)cot−1(23​)
  4. (D)tan⁡−1(3)\tan^{-1}(3)tan−1(3)

Correct answer: (C)

Step-by-step solution →
Q78·MathematicsSingle correctJEE Main 2021
Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin⁡−1(3x5)+sin⁡−1(4x5)=sin⁡−1x\sin^{-1}\left(\frac{3x}{5}\right)+\sin^{-1}\left(\frac{4x}{5}\right)=\sin^{-1}xsin−1(53x​)+sin−1(54x​)=sin−1x is equal to:
  1. (A)222
  2. (B)111
  3. (C)333
  4. (D)000

Correct answer: (C)

Step-by-step solution →
Q79·MathematicsSingle correctJEE Main 2021
If 0 < a, b < 1, and tan⁡−1a+tan⁡−1b=π4\tan^{-1} a + \tan^{-1} b = \frac{\pi}{4}tan−1a+tan−1b=4π​, then the value of (a+b)−(a2+b22)+(a3+b33)−(a4+b44)+…(a + b) - \left(\frac{a^{2} + b^{2}}{2}\right) + \left(\frac{a^{3} + b^{3}}{3}\right) - \left(\frac{a^{4} + b^{4}}{4}\right) + \ldots(a+b)−(2a2+b2​)+(3a3+b3​)−(4a4+b4​)+… is :
  1. (A)log⁡e2\log_{e} 2loge​2
  2. (B)log⁡e(e2)\log_{e}\left(\frac{e}{2}\right)loge​(2e​)
  3. (C)e
  4. (D)e2−1e^{2} - 1e2−1

Correct answer: (A)

Step-by-step solution →
Q80·MathematicsSingle correctJEE Main 2021
If sin⁡−1xa=cos⁡−1xb=tan⁡−1yc\frac{\sin^{-1}x}{a} = \frac{\cos^{-1}x}{b} = \frac{\tan^{-1}y}{c}asin−1x​=bcos−1x​=ctan−1y​ ; 0 < x < 1, then the value of cos⁡(πca+b)\cos\left(\frac{\pi c}{a+b}\right)cos(a+bπc​) is:
  1. (A)1−y22y\frac{1-y^{2}}{2y}2y1−y2​
  2. (B)1−y21+y2\frac{1-y^{2}}{1+y^{2}}1+y21−y2​
  3. (C)1−y21-y^{2}1−y2
  4. (D)1−y2yy\frac{1-y^{2}}{y\sqrt{y}}yy​1−y2​

Correct answer: (B)

Step-by-step solution →
Q81·MathematicsSingle correctJEE Main 2021
cosec [2cot–1^{–1}–1(5)+cos–1^{–1}–1 (54)] is equal to:
  1. (A)5675
  2. (B)5665
  3. (C)5633
  4. (D)3365

Correct answer: (B)

Step-by-step solution →
Q82·MathematicsSingle correctJEE Main 2021
A possible value of tan⁡(14sin⁡−1638)\tan\left(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\right)tan(41​sin−1863​​) is :
  1. (A)122\frac{1}{2\sqrt{2}}22​1​
  2. (B)17\frac{1}{\sqrt{7}}7​1​
  3. (C)7−1\sqrt{7}-17​−1
  4. (D)22−12\sqrt{2}-122​−1

Correct answer: (B)

Step-by-step solution →
Q83·MathematicsNumericalJEE Main 2021
lim⁡x→∞tan⁡{∑r=1ntan⁡−1(11+r+r2)}\lim_{x \to \infty} \tan\left\{ \sum_{r=1}^{n} \tan^{-1}\left(\frac{1}{1 + r + r^2}\right) \right\}limx→∞​tan{∑r=1n​tan−1(1+r+r21​)} is equal to ______

Correct answer: 1

Step-by-step solution →
Q84·MathematicsSingle correctJEE Main 2020
If S is the sum of the first 10 terms of the series tan⁡−1(13)+tan⁡−1(17)+tan⁡−1(113)+tan⁡−1(121)+....\tan^{-1}\left(\dfrac{1}{3}\right) + \tan^{-1}\left(\dfrac{1}{7}\right) + \tan^{-1}\left(\dfrac{1}{13}\right) + \tan^{-1}\left(\dfrac{1}{21}\right) + ....tan−1(31​)+tan−1(71​)+tan−1(131​)+tan−1(211​)+...., then tan(S) is equal to:
  1. (A)56\dfrac{5}{6}65​
  2. (B)1011\dfrac{10}{11}1110​
  3. (C)−65-\dfrac{6}{5}−56​
  4. (D)511\dfrac{5}{11}115​

Correct answer: (A)

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2020
2π−(sin⁡−145+sin⁡−1513+sin⁡−11665)2\pi - \left( \sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} + \sin^{-1}\frac{16}{65} \right)2π−(sin−154​+sin−1135​+sin−16516​) is equal to:
  1. (A)3π2\frac{3\pi}{2}23π​
  2. (B)5π4\frac{5\pi}{4}45π​
  3. (C)7π4\frac{7\pi}{4}47π​
  4. (D)π2\frac{\pi}{2}2π​

Correct answer: (A)

Step-by-step solution →
Q86·MathematicsSingle correctJEE Main 2020
Let f(x)=(sin⁡(tan⁡−1x)+sin⁡(cot⁡−1x))2−1,∣x∣>1f(x)=\left(\sin\left(\tan^{-1}x\right)+\sin\left(\cot^{-1}x\right)\right)^{2}-1, |x|>1f(x)=(sin(tan−1x)+sin(cot−1x))2−1,∣x∣>1. If dydx=12ddx(sin⁡−1(f(x)))\dfrac{dy}{dx}=\dfrac{1}{2}\dfrac{d}{dx}\left(\sin^{-1}(f(x))\right)dxdy​=21​dxd​(sin−1(f(x))) and y(3)=π6y\left(\sqrt{3}\right)=\dfrac{\pi}{6}y(3​)=6π​, then y(−3)y\left(-\sqrt{3}\right)y(−3​) is equal to
  1. (A)5π6\dfrac{5\pi}{6}65π​
  2. (B)π3\dfrac{\pi}{3}3π​
  3. (C)2π3\dfrac{2\pi}{3}32π​
  4. (D)−π6-\dfrac{\pi}{6}−6π​

Correct answer: (A)

Step-by-step solution →
Q87·MathematicsNumericalJEE Advanced 2019
The value of sec⁡−1(14∑k=010sec⁡(7π12+kπ2)sec⁡(7π12+(k+1)π2))\sec^{-1}\left( \dfrac{1}{4}\displaystyle\sum_{k=0}^{10} \sec\left(\dfrac{7\pi}{12} + \dfrac{k\pi}{2}\right)\sec\left(\dfrac{7\pi}{12} + \dfrac{(k+1)\pi}{2}\right) \right)sec−1(41​k=0∑10​sec(127π​+2kπ​)sec(127π​+2(k+1)π​)) in the interval [−π4, 3π4]\left[-\dfrac{\pi}{4},\ \dfrac{3\pi}{4}\right][−4π​, 43π​] equals ____

Correct answer: 0.00

Step-by-step solution →
Q88·MathematicsSingle correctJEE Main 2019
The value of sin⁡−1(1213)−sin⁡−1(35)\sin^{-1}\left(\dfrac{12}{13}\right) - \sin^{-1}\left(\dfrac{3}{5}\right)sin−1(1312​)−sin−1(53​) is equal to :
  1. (A)π−cos⁡−1(3365)\pi - \cos^{-1}\left(\dfrac{33}{65}\right)π−cos−1(6533​)
  2. (B)π−sin⁡−1(6365)\pi - \sin^{-1}\left(\dfrac{63}{65}\right)π−sin−1(6563​)
  3. (C)π2−cos⁡−1(965)\dfrac{\pi}{2} - \cos^{-1}\left(\dfrac{9}{65}\right)2π​−cos−1(659​)
  4. (D)π2−sin⁡−1(5665)\dfrac{\pi}{2} - \sin^{-1}\left(\dfrac{56}{65}\right)2π​−sin−1(6556​)

Correct answer: (D)

Step-by-step solution →
Q89·MathematicsSingle correctJEE Main 2019
If cos⁡−1x−cos⁡−1y2=α\cos^{-1}x - \cos^{-1}\dfrac{y}{2} = \alphacos−1x−cos−12y​=α, where -1 ≤ x ≤ 1, − 2 ≤ y ≤ 2, x≤y2x \le \dfrac{y}{2}x≤2y​, then for all x, y, 4x2−4xycos⁡α+y24x^{2} - 4xy \cos \alpha + y^{2}4x2−4xycosα+y2 is equal to
  1. (A)4sin⁡2α−2x2y24 \sin^{2}\alpha - 2x^{2}y^{2}4sin2α−2x2y2
  2. (B)4cos⁡2α+2x2y24 \cos^{2}\alpha + 2x^{2}y^{2}4cos2α+2x2y2
  3. (C)2sin⁡2α2 \sin^{2}\alpha2sin2α
  4. (D)4sin⁡2α4 \sin^{2}\alpha4sin2α

Correct answer: (D)

Step-by-step solution →
Q90·MathematicsSingle correctJEE Main 2019
If 2y=(cot⁡−1(3cos⁡x+sin⁡xcos⁡x−3sin⁡x))22y=\left(\cot^{-1}\left(\frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}\right)\right)^{2}2y=(cot−1(cosx−3​sinx3​cosx+sinx​))2, x∈(0,π2)x \in \left(0,\frac{\pi}{2}\right)x∈(0,2π​) then dydx\frac{dy}{dx}dxdy​ is equal to
  1. (A)π6−x\frac{\pi}{6}-x6π​−x
  2. (B)π3−x\frac{\pi}{3}-x3π​−x
  3. (C)x−π6x-\frac{\pi}{6}x−6π​
  4. (D)2x−π32x-\frac{\pi}{3}2x−3π​

Correct answer: (C)

Step-by-step solution →
Q91·MathematicsSingle correctJEE Main 2019
If α=cos⁡−1(35),β=tan⁡−1(13)\alpha=\cos^{-1}\left(\dfrac{3}{5}\right), \beta=\tan^{-1}\left(\dfrac{1}{3}\right)α=cos−1(53​),β=tan−1(31​), where 0<α,β<π20<\alpha,\beta<\dfrac{\pi}{2}0<α,β<2π​, then α−β\alpha-\betaα−β is equal to:
  1. (A)sin⁡−1(9510)\sin^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)sin−1(510​9​)
  2. (B)cos⁡−1(9510)\cos^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)cos−1(510​9​)
  3. (C)tan⁡−1(9510)\tan^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)tan−1(510​9​)
  4. (D)tan⁡−1(914)\tan^{-1}\left(\dfrac{9}{14}\right)tan−1(149​)

Correct answer: (A)

Step-by-step solution →
Q92·MathematicsSingle correctJEE Main 2019
Considering only the principal values of inverse functions, the set A={x≥0:tan⁡−1(2x)+tan⁡−1(3x)=π4}A=\left\{x\geq 0: \tan^{-1}(2x)+\tan^{-1}(3x)=\frac{\pi}{4}\right\}A={x≥0:tan−1(2x)+tan−1(3x)=4π​}
  1. (A)contains two elements
  2. (B)contains more than two elements
  3. (C)is a singleton
  4. (D)is an empty set

Correct answer: (C)

Step-by-step solution →
Q93·MathematicsSingle correctJEE Main 2019
All x satisfying the inequality (cot⁡−1x)2−7(cot⁡−1x)+10>0\left(\cot^{-1}x\right)^{2}-7\left(\cot^{-1}x\right)+10>0(cot−1x)2−7(cot−1x)+10>0, lie in the interval:
  1. (A)(−∞,cot⁡5)∪(cot⁡4,cot⁡2)(-\infty,\cot 5)\cup(\cot 4,\cot 2)(−∞,cot5)∪(cot4,cot2)
  2. (B)(cot⁡2,∞)(\cot 2,\infty)(cot2,∞)
  3. (C)(−∞,cot⁡5)∪(cot⁡2,∞)(-\infty,\cot 5)\cup(\cot 2,\infty)(−∞,cot5)∪(cot2,∞)
  4. (D)(cot⁡5,cot⁡4)(\cot 5,\cot 4)(cot5,cot4)

Correct answer: (B)

Step-by-step solution →
Q94·MathematicsSingle correctJEE Main 2019
If cos⁡−1(23x)+cos⁡−1(34x)=π2\cos^{-1}\left(\dfrac{2}{3x}\right) + \cos^{-1}\left(\dfrac{3}{4x}\right) = \dfrac{\pi}{2}cos−1(3x2​)+cos−1(4x3​)=2π​, x>34x > \dfrac{3}{4}x>43​ then x is equal to:
  1. (A)14512\dfrac{\sqrt{145}}{12}12145​​
  2. (B)14510\dfrac{\sqrt{145}}{10}10145​​
  3. (C)14612\dfrac{\sqrt{146}}{12}12146​​
  4. (D)14511\dfrac{\sqrt{145}}{11}11145​​

Correct answer: (A)

Step-by-step solution →
Q95·MathematicsSingle correctJEE Main 2019
If x=sin⁡−1(sin⁡10)x=\sin^{-1}(\sin 10)x=sin−1(sin10) and y=cos⁡−1(cos⁡10)y=\cos^{-1}(\cos 10)y=cos−1(cos10), then y−xy-xy−x is equal to:
  1. (A)π\piπ
  2. (B)7π7\pi7π
  3. (C)0
  4. (D)10

Correct answer: (A)

Step-by-step solution →
Q96·MathematicsNumericalJEE Advanced 2018
The number of real solutions of the equation sin⁡−1(∑i=1∞xi+1−x∑i=1∞(x2)i)=π2−cos⁡−1(∑i=1∞(−x2)i−∑i=1∞(−x)i)\sin^{-1}\left(\sum_{i=1}^{\infty} x^{i+1} - x \sum_{i=1}^{\infty}\left(\frac{x}{2}\right)^{i}\right) = \frac{\pi}{2} - \cos^{-1}\left(\sum_{i=1}^{\infty}\left(-\frac{x}{2}\right)^{i} - \sum_{i=1}^{\infty}(-x)^{i}\right)sin−1(∑i=1∞​xi+1−x∑i=1∞​(2x​)i)=2π​−cos−1(∑i=1∞​(−2x​)i−∑i=1∞​(−x)i) lying in the interval (−12,12)\left(-\frac{1}{2}, \frac{1}{2}\right)(−21​,21​) is ______ . (Here, the inverse trigonometric functions sin⁡−1x\sin^{-1}xsin−1x and cos⁡−1x\cos^{-1}xcos−1x assume values in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right][−2π​,2π​] and [0,π][0, \pi][0,π], respectively.)

Correct answer: 2

Step-by-step solution →
Q97·MathematicsMultiple correctJEE Advanced 2015
If α=3sin⁡−1(611)\alpha = 3\sin^{-1}\left(\dfrac{6}{11}\right)α=3sin−1(116​) and β=3cos⁡−1(49)\beta = 3\cos^{-1}\left(\dfrac{4}{9}\right)β=3cos−1(94​), where the inverse trigonometric functions take only the principal values, then the correct option(s) is(are)
  1. (A)cos⁡β>0\cos\beta > 0cosβ>0
  2. (B)sin⁡β<0\sin\beta < 0sinβ<0
  3. (C)cos⁡(α+β)>0\cos(\alpha + \beta) > 0cos(α+β)>0
  4. (D)cos⁡α<0\cos\alpha < 0cosα<0

Correct answer: (B), (C), (D)

Step-by-step solution →
Q98·MathematicsSingle correctJEE Advanced 2014
Match the following:
List – IList – II
P.Let y(x)=cos⁡(3cos⁡−1x)y(x) = \cos(3\cos^{-1} x)y(x)=cos(3cos−1x), x∈[−1,1]x \in [-1, 1]x∈[−1,1], x≠±32x \ne \pm\frac{\sqrt{3}}{2}x=±23​​. Then 1y(x){(x2−1)d2y(x)dx2+xdy(x)dx}\frac{1}{y(x)}\left\{(x^{2}-1)\frac{d^{2}y(x)}{dx^{2}} + x\frac{dy(x)}{dx}\right\}y(x)1​{(x2−1)dx2d2y(x)​+xdxdy(x)​} equals1.1
Q.Let A1,A2,…,AnA_{1}, A_{2}, \ldots, A_{n}A1​,A2​,…,An​ (n>2)(n > 2)(n>2) be the vertices of a regular polygon of nnn sides with its centre at the origin. Let ak→\overrightarrow{a_{k}}ak​​ be the position vector of the point AkA_{k}Ak​, k=1,2,…,nk = 1, 2, \ldots, nk=1,2,…,n. If ∣∑k=1n−1(ak→×ak+1→)∣=∣∑k=1n−1(ak→⋅ak+1→)∣\left|\sum_{k=1}^{n-1}\left(\overrightarrow{a_{k}} \times \overrightarrow{a_{k+1}}\right)\right| = \left|\sum_{k=1}^{n-1}\left(\overrightarrow{a_{k}} \cdot \overrightarrow{a_{k+1}}\right)\right|​∑k=1n−1​(ak​​×ak+1​​)​=​∑k=1n−1​(ak​​⋅ak+1​​)​, then the minimum value of nnn is2.2
R.If the normal from the point P(h,1)P(h, 1)P(h,1) on the ellipse x26+y23=1\frac{x^{2}}{6} + \frac{y^{2}}{3} = 16x2​+3y2​=1 is perpendicular to the line x+y=8x + y = 8x+y=8, then the value of hhh is3.8
S.Number of positive solutions satisfying the equation tan⁡−1(12x+1)+tan⁡−1(14x+1)=tan⁡−1(2x2)\tan^{-1}\left(\frac{1}{2x+1}\right) + \tan^{-1}\left(\frac{1}{4x+1}\right) = \tan^{-1}\left(\frac{2}{x^{2}}\right)tan−1(2x+11​)+tan−1(4x+11​)=tan−1(x22​) is4.9
  1. (A)P-4, Q-3, R-2, S-1
  2. (B)P-2, Q-4, R-3, S-1
  3. (C)P-4, Q-3, R-1, S-2
  4. (D)P-2, Q-4, R-1, S-3

Correct answer: (A)

Step-by-step solution →
Q99·MathematicsIntegerJEE Advanced 2014
Let f:[0,4π]→[0,π]f: [0, 4\pi] \to [0, \pi]f:[0,4π]→[0,π] be defined by f(x)=cos⁡−1(cos⁡x)f(x) = \cos^{-1}(\cos x)f(x)=cos−1(cosx). The number of points x∈[0,4π]x \in [0, 4\pi]x∈[0,4π] satisfying the equation f(x)=10−x10f(x) = \frac{10 - x}{10}f(x)=1010−x​ is __________

Correct answer: 3

Step-by-step solution →
Q100·MathematicsSingle correctJEE Advanced 2013
Match List I with List II and select the correct answer using the code given below the lists :
List-IList-II
P.(1y2(cos⁡(tan⁡−1y)+ysin⁡(tan⁡−1y)cot⁡(sin⁡−1y)+tan⁡(sin⁡−1y))2+y4)1/2\left(\frac{1}{y^{2}}\left(\frac{\cos\left(\tan^{-1} y\right) + y\sin\left(\tan^{-1} y\right)}{\cot\left(\sin^{-1} y\right) + \tan\left(\sin^{-1} y\right)}\right)^{2} + y^{4}\right)^{1/2}(y21​(cot(sin−1y)+tan(sin−1y)cos(tan−1y)+ysin(tan−1y)​)2+y4)1/2 takes value1.1253\frac{1}{2}\sqrt{\frac{5}{3}}21​35​​
Q.If cos⁡x+cos⁡y+cos⁡z=0=sin⁡x+sin⁡y+sin⁡z\cos x + \cos y + \cos z = 0 = \sin x + \sin y + \sin zcosx+cosy+cosz=0=sinx+siny+sinz then possible value of cos⁡x−y2\cos\frac{x - y}{2}cos2x−y​ is2.2\sqrt{2}2​
R.If cos⁡(π4−x)cos⁡2x+sin⁡xsin⁡2xsec⁡x=cos⁡xsin⁡2xsec⁡x+cos⁡(π4+x)cos⁡2x\cos\left(\frac{\pi}{4} - x\right)\cos 2x + \sin x \sin 2x \sec x = \cos x \sin 2x \sec x + \cos\left(\frac{\pi}{4} + x\right)\cos 2xcos(4π​−x)cos2x+sinxsin2xsecx=cosxsin2xsecx+cos(4π​+x)cos2x then possible value of sec⁡x\sec xsecx is3.12\frac{1}{2}21​
S.If cot⁡(sin⁡−11−x2)=sin⁡(tan⁡−1(x6))\cot\left(\sin^{-1}\sqrt{1 - x^{2}}\right) = \sin\left(\tan^{-1}\left(x\sqrt{6}\right)\right)cot(sin−11−x2​)=sin(tan−1(x6​)), x≠0x \neq 0x=0, then possible value of xxx is4.111
  1. (A)P-4, Q-3, R-1, S-2
  2. (B)P-4, Q-3, R-2, S-1
  3. (C)P-3, Q-4, R-2, S-1
  4. (D)P-3, Q-4, R-1, S-2

Correct answer: (B)

Step-by-step solution →
Q101·MathematicsSingle correctJEE Advanced 2013
The value of cot⁡(∑n=123cot⁡−1(1+∑k=1n2k))\cot\left(\sum_{n=1}^{23}\cot^{-1}\left(1+\sum_{k=1}^{n}2k\right)\right)cot(∑n=123​cot−1(1+∑k=1n​2k)) is
  1. (A)2325\frac{23}{25}2523​
  2. (B)2523\frac{25}{23}2325​
  3. (C)2324\frac{23}{24}2423​
  4. (D)2423\frac{24}{23}2324​

Correct answer: (B)

Step-by-step solution →

Inverse Trigonometric Functions — frequently asked

How many questions from Inverse Trigonometric Functions appear in JEE?

Inverse Trigonometric Functions has appeared in 93 of the last 186 JEE Main and JEE Advanced papers — about 50% of them — contributing 101 questions in total across those papers.

Is Inverse Trigonometric Functions an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 50% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Inverse Trigonometric Functions questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226

All 26 Mathematics chapters →

Practise Inverse Trigonometric Functions until it stops costing you marks.

Build a timed test from these 101 questions in one click. Jarvis marks it, names the specific misconception behind each wrong answer, and brings the ones you failed back at the right interval.

Practise Inverse Trigonometric Functions freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS