Inverse Trigonometric Functions
100 JEE Maths previous year questions on Inverse Trigonometric Functions — options free on every question; 10 include the answer & explanation free, the rest unlock with PYQ Pass.
Let \([x]\) denote the greatest integer less than or equal to \(x\). Then the domain of \(f(x)=\sec ^{-1}(2[x]+1)\) is :
[JEE Main 2025, 28 Jan (Shift 2)]
\((-\infty ,\infty )\)
Domain of \({\text{sec}}^{-1}x\) is \(\mathrm{x}\in (-\infty ,-1]\cup [1,\infty )\)
\(2[x]+1 \leq-1\) or \(2[x]+1 \geq 1\)
\(\Rightarrow[x] \leq-1\) or \([x] \geq 0\)
\(\Rightarrow \mathrm{x} \in(-\infty, 0)\) or \(x \in[0, \infty)\)
\(\Rightarrow x \in(-\infty, \infty)\)
Using the principal values of the inverse trigonometric functions the sum of the maximum and the minimum values of \(16\left({\left({\sec }^{-1}x\right)}^{2}+{\left({\csc }^{-1}x\right)}^{2}\right)\) is:
[JEE Main 2025, 22 Jan (Shift 1)]
\(22 \pi^2\)
\(16\left({\left({\sec }^{-1}x\right)}^{2}+{\left({\csc }^{-1}x\right)}^{2}\right)............\text{(i)}\\ \text{we know that,}se{c}^{-1}x+\cos e{c}^{-1}x=\frac{\pi }{2}\\ \text{and}se{c}^{-1}x\in \left[0,\pi \right]-\left\{\frac{\pi }{2}\right\}\\ \text{now, put in (i), we get}\\ 16\left({\left({\sec }^{-1}x\right)}^{2}+{\left(\frac{\pi }{2}-se{c}^{-1}x\right)}^{2}\right)\\ letse{c}^{-1}x=y\\ 16\left(2{y}^{2}-\pi y+\frac{{\pi }^{2}}{4}\right)=0\\ \text{Now}\text{, }\text{it}\text{ }\text{is}\text{ }\text{maximum}\text{ }\text{at}y=\pi \\ \text{maximum}\text{ }\text{value}=16[2{\pi }^{2}-{\pi }^{2}+\frac{{\pi }^{2}}{4}]\\ =20{\pi }^{2}\\ \text{and}\text{ }\text{minimum}\text{ }\text{at}y=\frac{\pi }{4}\\ (y=-\frac{b}{2a}\text{for the quadratic in}y)\\ \text{minimum}\text{ }\text{value}=16[\frac{2\times {\pi }^{2}}{16}-\frac{{\pi }^{2}}{4}+\frac{{\pi }^{2}}{4}]\\ =2{\pi }^{2}\\ \text{And}\text{ }\text{sum}\text{ }\text{of}\text{ }\text{maximum}\text{ \& }\text{minimum}\text{ value}=\\ 20{\pi }^{2}+2{\pi }^{2}=22{\pi }^{2}\)
Given that the inverse trigonometric function assumes principal values only. Let \(x\), \(y\) be any two real numbers in \([-1,1]\) such that \({\cos }^{-1}x-{\sin }^{-1}y=\alpha ,\frac{-\pi }{2}\leq \alpha \leq \pi\). Then, the minimum value of \({x}^{2}+{y}^{2}+2xy\sin \alpha\) is
[JEE Main 2024, 4 Apr (Shift 2)]
\(0\)
Let \({\cos }^{-1}x=A\) and \({\sin }^{-1}y=B\). Then \(x=\cos A\) and \(y=\sin B\).
given condition is \(A-B=\alpha\).
The expression is \(E={x}^{2}+{y}^{2}+2xy\sin \alpha\).
Substitute \(x,y\), and \(\alpha\) :
\(E={\cos }^{2}A+{\sin }^{2}B+2\cos A\sin B\sin (A-B)\)
Using the identity \(\sin (A-B)=\sin A\cos B-\cos A\sin B\) :
\(E={\cos }^{2}A+{\sin }^{2}B+2\cos A\sin B(\sin A\cos B-\cos A\sin B)\)
\(E={\cos }^{2}A+{\sin }^{2}B+2\sin A\cos A\sin B\cos B-2{\cos }^{2}A{\sin }^{2}B\)
\(E={\cos }^{2}A\left(1-{\sin }^{2}B\right)+{\sin }^{2}B\left(1-{\cos }^{2}A\right)+2\sin A\cos A\sin B\cos B\)
\(E={\cos }^{2}A{\cos }^{2}B+{\sin }^{2}A{\sin }^{2}B+2\sin A\sin B\cos A\cos B\\ E=(\cos A\cos B+\sin A\sin B{)}^{2}=(\cos (A-B){)}^{2}={\cos }^{2}\alpha\)
The minimum value of \({\cos }^{2}\alpha \text{ is }0\)
At \(\alpha=\frac{\pi}{2}\)
Let \([\cdot ]\) denote the greatest integer function. If the domain of the function \(f\left(x\right)={\cos }^{-1}\left(\frac{4x+2[x]}{3}\right)\) is\([\alpha ,\beta ]\)then \(12(\alpha +\beta )\) is equal to:
[JEE Main 2026, 4 Apr (Shift 1)]
\(6\)
For \(f(x)=\cos^{-1}\left(\dfrac{4x+2[x]}{3}\right)\) to be defined, we need \(-1\leq \dfrac{4x+2[x]}{3}\leq 1\).
So \(-3\leq 4x+2[x]\leq 3\).
Let \([x]=n\), where \(n\in \mathbb{Z}\). Then \(n\leq x The inequality becomes \(-3\leq 4x+2n\leq 3\), hence \(\dfrac{-3-2n}{4}\leq x\leq \dfrac{3-2n}{4}\). Now check possible integer values of \(n\). For \(n=-1\), \(-1\leq x<0\) and \(-\dfrac{1}{4}\leq x\leq \dfrac{5}{4}\), giving \(-\dfrac{1}{4}\leq x<0\). For \(n=0\), \(0\leq x<1\) and \(-\dfrac{3}{4}\leq x\leq \dfrac{3}{4}\), giving \(0\leq x\leq \dfrac{3}{4}\). No other integer value of \(n\) gives a common interval. Thus the domain is \(\left[-\dfrac{1}{4},0\right)\cup\left[0,\dfrac{3}{4}\right]=\left[-\dfrac{1}{4},\dfrac{3}{4}\right]\). So \(\alpha=-\dfrac{1}{4}\) and \(\beta=\dfrac{3}{4}\). Therefore \(12(\alpha+\beta)=12\left(-\dfrac{1}{4}+\dfrac{3}{4}\right)=12\cdot\dfrac{1}{2}=6\).
If \(f(x)=16\left(\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\right)\) then the sum of max. and min. value of \(f(x)\) is (22 Jan, Shift I, Memory Based)
\(22{\pi }^{2}\)
\(f(x)=16\left({\left({\sec }^{-1}x\right)}^{2}+{\left({\csc }^{-1}x\right)}^{2}\right)\\ \Rightarrow f(x)=16\left({\left({\sec }^{-1}x\right)}^{2}+{\left(\frac{\pi }{2}-{\sec }^{-1}x\right)}^{2}\right)\\ \Rightarrow f(x)=32\left({\left({\sec }^{-1}x-\frac{\pi }{4}\right)}^{2}+\frac{{\pi }^{2}}{16}\right)\\ \Rightarrow f{(x)}_{min}=32\times \frac{{\pi }^{2}}{16}=2{\pi }^{2}\\ \Rightarrow \mathrm{f}{(\mathrm{x})}_{\max }=32\left({\left(\pi -\frac{\pi }{4}\right)}^{2}+\frac{{\pi }^{2}}{16}\right)=20{\pi }^{2}\\ \Rightarrow \mathrm{f}{(\mathrm{x})}_{\min }+\mathrm{f}{(\mathrm{x})}_{\max }=22{\pi }^{2}\)
Using the principal values of the inverse trigonometric functions the sum of the maximum and the minimum values of \(16\left({\left({\sec }^{-1}x\right)}^{2}+{\left({\csc }^{-1}x\right)}^{2}\right)\) is:
[JEE Main 2025, 22 Jan (Shift 1)]
\(22 \pi^2\)
\(16\left({\left({\sec }^{-1}x\right)}^{2}+{\left({\csc }^{-1}x\right)}^{2}\right)............\text{(i)}\\ \text{we know that,}se{c}^{-1}x+\cos e{c}^{-1}x=\frac{\pi }{2}\\ \text{and}se{c}^{-1}x\in \left[0,\pi \right]-\left\{\frac{\pi }{2}\right\}\\ \text{now, put in (i), we get}\\ 16\left({\left({\sec }^{-1}x\right)}^{2}+{\left(\frac{\pi }{2}-se{c}^{-1}x\right)}^{2}\right)\\ letse{c}^{-1}x=y\\ 16\left(2{y}^{2}-\pi y+\frac{{\pi }^{2}}{4}\right)=0\\ \text{Now}\text{, }\text{it}\text{ }\text{is}\text{ }\text{maximum}\text{ }\text{at}y=\pi \\ \text{maximum}\text{ }\text{value}=16[2{\pi }^{2}-{\pi }^{2}+\frac{{\pi }^{2}}{4}]\\ =20{\pi }^{2}\\ \text{and}\text{ }\text{minimum}\text{ }\text{at}y=\frac{\pi }{4}\\ (y=-\frac{b}{2a}\text{for the quadratic in}y)\\ \text{minimum}\text{ }\text{value}=16[\frac{2\times {\pi }^{2}}{16}-\frac{{\pi }^{2}}{4}+\frac{{\pi }^{2}}{4}]\\ =2{\pi }^{2}\\ \text{And}\text{ }\text{sum}\text{ }\text{of}\text{ }\text{maximum}\text{ \& }\text{minimum}\text{ value}=\\ 20{\pi }^{2}+2{\pi }^{2}=22{\pi }^{2}\)
Let \(0<\alpha <1,\beta =\frac{1}{3\alpha }\) and \({\tan }^{-1}(1-\alpha )+{\tan }^{-1}(1-\beta )=\frac{\pi }{4}\). Then \(6(\alpha +\beta )\) is equal to:
[JEE Main 2026, 6 Apr (Shift 1)]
\(7\)
Let \(A=\tan^{-1}(1-\alpha)\) and \(B=\tan^{-1}(1-\beta)\).
Given \(A+B=\frac{\pi}{4}\), so \(\tan(A+B)=1\).
\(\frac{(1-\alpha)+(1-\beta)}{1-(1-\alpha)(1-\beta)}=1\)
\(\frac{2-\alpha-\beta}{\alpha+\beta-\alpha\beta}=1\)
\(2-\alpha-\beta=\alpha+\beta-\alpha\beta\)
\(2=2\alpha+2\beta-\alpha\beta\)
Using \(\beta=\frac{1}{3\alpha}\),
\(2=2\alpha+\frac{2}{3\alpha}-\frac{1}{3}\)
Multiplying by \(3\alpha\),
\(6\alpha=6\alpha^2+2-\alpha\)
\(6\alpha^2-7\alpha+2=0\)
\((3\alpha-2)(2\alpha-1)=0\)
\(\alpha=\frac{2}{3}\) or \(\alpha=\frac{1}{2}\)
If \(\alpha=\frac{2}{3}\), then \(\beta=\frac{1}{2}\);
if \(\alpha=\frac{1}{2}\), then \(\beta=\frac{2}{3}\).
So, \(\alpha+\beta=\frac{7}{6}\)
\(6(\alpha+\beta)=6\cdot\frac{7}{6}=7\)
If \(\cot \left(\cos ^{-1} x\right)=\sec \left(\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right)\), then:
[JEE Main 2024]
\(\frac{b}{\sqrt{2 b^2-a^2}}\)
\(\text{ Given, }\cot \left({\cos }^{-1}x\right)=\sec \left({\tan }^{-1}\frac{a}{\sqrt{{b}^{2}-{a}^{2}}}\right)\\ \Rightarrow cot\left({\cot }^{-1}\left(\frac{x}{\sqrt{1-{x}^{2}}}\right)\right)=\sec \left({\sec }^{-1}\frac{b}{\sqrt{{b}^{2}-{a}^{2}}}\right)\\ \Rightarrow \frac{x}{\sqrt{1-{x}^{2}}}=\frac{b}{\sqrt{{b}^{2}-{a}^{2}}}\\ \Rightarrow \frac{b}{\sqrt{2{b}^{2}-{a}^{2}}}=x\)
If \(\cot \left(\cos ^{-1} x\right)=\sec \left(\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right)\), then:
[JEE Main 2024]
\(\frac{b}{\sqrt{2 b^2-a^2}}\)
\(\text{ Given, }\cot \left({\cos }^{-1}x\right)=\sec \left({\tan }^{-1}\frac{a}{\sqrt{{b}^{2}-{a}^{2}}}\right)\\ \Rightarrow cot\left({\cot }^{-1}\left(\frac{x}{\sqrt{1-{x}^{2}}}\right)\right)=\sec \left({\sec }^{-1}\frac{b}{\sqrt{{b}^{2}-{a}^{2}}}\right)\\ \Rightarrow \frac{x}{\sqrt{1-{x}^{2}}}=\frac{b}{\sqrt{{b}^{2}-{a}^{2}}}\\ \Rightarrow \frac{b}{\sqrt{2{b}^{2}-{a}^{2}}}=x\)
For \(\alpha, \beta, \gamma \neq 0\), if \(\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi\) and \((\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta\), then \(\gamma\) equals
[JEE Main 2024, 31 Jan (Shift 1)]
\(\frac{\sqrt{3}}{2}\)
\(A=\sin^{-1}\alpha,\quad B=\sin^{-1}\beta,\quad C=\sin^{-1}\gamma\)
\(A+B+C=\pi\)
\(\gamma=\sin C=\sin(A+B)\)
\(=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}\)
\((\alpha+\beta+\gamma)(\alpha+\beta-\gamma)=3\alpha\beta\)
\((\alpha+\beta)^2-\gamma^2=3\alpha\beta\)
\(\gamma=\sin(A+B)\)
\((\sin A+\sin B)^2-\sin^2(A+B)=3\sin A\sin B\)
\(\sin(A+B)=\sin A\cos B+\cos A\sin B\)
\(\sin A\sin B\bigl(1+\cos(A+B)\bigr)\)
\(\sin A\sin B\bigl(1+\cos(A+B)\bigr)=3\sin A\sin B\)
Since \(\alpha,\beta\ne0\)
\(\sin A\sin B\ne0\)
\(1+\cos(A+B)=3\)
\(\cos(A+B)=\dfrac12\)
\(A+B=\pi-C\)
\(\cos(A+B)=\cos(\pi-C)=-\cos C\)
\(-\cos C=\dfrac12\)
\(\cos C=-\dfrac12\)
\(C\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\)
\((\alpha+\beta)^2-\gamma^2=3\alpha\beta\)
\(\alpha^2+\beta^2-\alpha\beta=\gamma^2\)
\(\gamma=\sin(A+B)\)
\(\sin^2(A+B)=\alpha^2+\beta^2-\alpha\beta\)
\(\sin^2(A+B)=(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})^2\)
\(2\sqrt{(1-\alpha^2)(1-\beta^2)}=1\)
\((1-\alpha^2)(1-\beta^2)=\dfrac14\)
\(\cos(A+B)=\sqrt{1-\alpha^2}\sqrt{1-\beta^2}-\alpha\beta\)
\(=\dfrac12-\alpha\beta\)
\(C=\pi-(A+B)\)
\(\gamma=\sin(A+B)\)
\(\gamma^2=\alpha^2+\beta^2-\alpha\beta\)
\(\alpha^2+\beta^2=1+\alpha^2\beta^2-\dfrac14\)
\(\gamma^2=\dfrac34\)
\(\sin^{-1}\alpha+\sin^{-1}\beta+\sin^{-1}\gamma=\pi\)
all three quantities are positive, so \(\gamma>0\)
\(\gamma=\dfrac{\sqrt3}{2}\).
If \(\alpha>\beta>\gamma>0\), then the expression \({\cot }^{-1}\left\{\beta +\frac{\left(1+{\beta }^{2}\right)}{(\alpha -\beta )}\right\}\)\(+{\cot }^{-1}\left\{\gamma +\frac{\left(1+{\gamma }^{2}\right)}{(\beta -\gamma )}\right\}\)\(+{\cot }^{-1}\left\{\alpha +\frac{\left(1+{\alpha }^{2}\right)}{(\gamma -\alpha )}\right\}\) is equal to :
[JEE Main 2025, 24 Jan (Shift 2)]
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\(\text{ Find domain of }{\sec }^{-1}(2[x]+1)\text{,}\)
(where [ ] denotes greatest integer function)
[JEE Main 2025]
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Let \([\cdot]\) denote that greatest integer function. If the domain of the function \(f(x)=\sin ^{-1}\left(\frac{x+[x]}{3}\right)\) is \([\alpha, \beta)\), then \(\alpha^2+\beta^2\) is equal to:
[JEE Main 2026, 6 Apr (Shift 1)]
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\(\text{ Find domain of }{\sec }^{-1}(2[x]+1)\text{,}\)
(where [ ] denotes greatest integer function)
[JEE Main 2025]
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Considering only the principal values of inverse trigonometric functions, the number of positive real values of \(x\) satisfying \(\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4}\) is :
[JEE Main 2024, 27 Jan (Shift 2)]
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If \(\alpha\) and \(\beta\) are real numbers such that \(\sec ^2\left(\tan ^{-1}(\alpha)\right)+\operatorname{cosec}^2\left(\cot ^{-1}(\beta)\right)=36\) and \(\alpha+\beta=8\), then \(\left(\alpha^2+\beta\right)\) is \((\alpha>\beta)\) (24 Jan, Shift I, Memory Based)
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\(\text{ If }{\cos }^{-1}x=\pi +{\sin }^{-1}x+{\sin }^{-1}(2x-1)\text{,}\\ \text{then find the sum of all values of ' }x\text{ '. }\)
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If the domain of the function \(f\left(x\right)={\sin }^{-1}\left(\frac{x-1}{2x+3}\right)\) is \(R-(\alpha ,\beta )\), then \(12\alpha \beta\) is equal to :
[JEE Main 2024, 9 Apr (Shift 1)]
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Value of \(\cos ^{-1}\left[\frac{12}{13} \cos x+\frac{5}{13} \sin x\right]\) is \(\left(x \in\left[\frac{\pi}{2}, \frac{3\pi}{4}\right]\right)\)
[JEE Main 2025]
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Considering only the principal values of the inverse trigonometric functions, the value of \(\tan \left({\sin }^{-1}\left(\frac{3}{5}\right)-2{\cos }^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)\) is
[JEE Advanced 2024]
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Let \([x]\) denote the greatest integer less than or equal to \(x\). Then the domain of \(f(x)=\sec ^{-1}(2[x]+1)\) is :
[JEE Main 2025, 28 Jan (Shift 2)]
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If the domain of the function \({\sin }^{-1}\left(\frac{3x-22}{2x-19}\right)+{\log }_{e}\left(\frac{3{x}^{2}-8x+5}{{x}^{2}-3x-10}\right)\) is \((\alpha ,\beta ]\), then \(3\alpha +10\beta\) is equal to:
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If \(\sin \left({\tan }^{-1}(x\sqrt{2})\right)=\cot \left({\sin }^{-1}\sqrt{1-{x}^{2}}\right),x\in \left(0,1\right)\), then the value of \(x\) is:
[JEE Main 2026, 6 Apr (Shift 2)]
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Let \(x=\frac{m}{n}\) ( \(m, n\) are co-prime natural numbers) be a solution of the equation \(\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}\) and let \(\alpha, \beta(\alpha>\beta)\) be the roots of the equation \(m x^2-n x-m+\) \(n=0\). Then the point \((\alpha, \beta)\) lies on the line
[JEE Main 2024, 29 Jan (Shift 2)]
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Let \(x=\frac{m}{n}\) ( \(m, n\) are co-prime natural numbers) be a solution of the equation \(\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}\) and let \(\alpha, \beta(\alpha>\beta)\) be the roots of the equation \(m x^2-n x-m+\) \(n=0\). Then the point \((\alpha, \beta)\) lies on the line
[JEE Main 2024, 29 Jan (Shift 2)]
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\(\cos \left({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{33}{65}\right)\text{ is equal to: }\) (28 Jan, Shift I, Memory Based)
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Value of \(\cos ^{-1}\left[\frac{12}{13} \cos x+\frac{5}{13} \sin x\right]\) is \(\left(x \in\left[\frac{\pi}{2}, \frac{3\pi}{4}\right]\right)\)
[JEE Main 2025]
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The value of \({\cot }^{-1}\left(\frac{\sqrt{1+{\tan }^{2}(2)}-1}{\tan (2)}\right)-{\cot }^{-1}\) \(\left(\frac{\sqrt{1+{\tan }^{2}\left(\frac{1}{2}\right)}+1}{\tan \left(\frac{1}{2}\right)}\right)\) is equal to
[JEE Main 2025, 8 Apr (Shift 1)]
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The sum of the infinite series \({\cot }^{-1}\left(\frac{7}{4}\right)+{\cot }^{-1}\left(\frac{19}{4}\right)+{\cot }^{-1}\left(\frac{39}{4}\right)+{\cot }^{-1}\left(\frac{67}{4}\right)+\ldots .\). is :-
[JEE Main 2025, 4 Apr (Shift 2)]
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If \(\alpha>\beta>\gamma>0\), then the expression \({\cot }^{-1}\left\{\beta +\frac{\left(1+{\beta }^{2}\right)}{(\alpha -\beta )}\right\}\)\(+{\cot }^{-1}\left\{\gamma +\frac{\left(1+{\gamma }^{2}\right)}{(\beta -\gamma )}\right\}\)\(+{\cot }^{-1}\left\{\alpha +\frac{\left(1+{\alpha }^{2}\right)}{(\gamma -\alpha )}\right\}\) is equal to :
[JEE Main 2025, 24 Jan (Shift 2)]
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The total number of real solutions of the equation\(\ \theta=\tan ^{-1}(2 \tan \theta)-\frac{1}{2} \sin ^{-1}\left(\frac{6 \tan \theta}{9+\tan ^2 \theta}\right) \) is (Here, the inverse trigonometric functions \(\sin ^{-1} x\) and \(\tan ^{-1} x\) assume values in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), respectively.
[JEE Advanced 2025]
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\(\text { If } \alpha>\beta>\gamma>0 \text {, then find } \cot ^{-1}\left(\frac{1+\alpha \beta}{\alpha-\beta}\right)+\cot ^{-1}\left(\frac{1+\beta \gamma}{\beta-\gamma}\right)+\cot ^{-1}\left(\frac{1+\gamma \alpha}{\gamma-\alpha}\right)\) (24 Jan, Shift II, Memory Based)
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Consider the principal values of inverse trigonometric functions, the value of the expression \(\tan \left(2{\sin }^{−1}\left(\frac{2}{\sqrt{13}}\right)−2{\cos }^{−1}\left(\frac{3}{\sqrt{10}}\right)\right)\) is equal to:
[JEE Main 2026, 28 Jan (Shift 2)]
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If \(a={\sin }^{-1}(\sin (5))\) and \(b={\cos }^{-1}(\cos (5))\), then \({a}^{2}+{b}^{2}\) is equal to
[JEE Main 2024, 31 Jan (Shift 2)]
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Let \(\alpha =3{\sin }^{-1}\left(\frac{6}{11}\right)\) and \(\beta =3{\cos }^{-1}\left(\frac{4}{9}\right)\), where inverse trigonometric functions take only the principal values.
Given below are two statements:
Statement I: \(\cos (\alpha +\beta )>0\).
Statement II: \(\cos (\alpha )<0\).
In the light of the above statements, choose the correct answer from the options given below:
[JEE Main 2026, 8 Apr (Shift 2)]
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Let \(x=\frac{m}{n}\) ( \(m, n\) are co-prime natural numbers) be a solution of the equation \(\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}\) and let \(\alpha, \beta(\alpha>\beta)\) be the roots of the equation \(m x^2-n x-m+\) \(n=0\). Then the point \((\alpha, \beta)\) lies on the line
[JEE Main 2024, 29 Jan (Shift 2)]
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Using the principal values of the inverse trigonometric functions the sum of the maximum and the minimum values of \(16\left({\left({\sec }^{-1}x\right)}^{2}+{\left({\csc }^{-1}x\right)}^{2}\right)\) is:
[JEE Main 2025, 22 Jan (Shift 1)]
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If \(\frac{\pi }{2}\leq x\leq \frac{3\pi }{4}\) , then \({\cos }^{-1}\left(\frac{12}{13}\cos x+\frac{5}{13}\sin x\right)\) is equal to
[JEE Main 2025, 23 Jan (Shift 1)]
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Considering only the principal values of the inverse trigonometric functions, the value of \({\text{cot}}^{−1}\left(\text{cot}\left(−11\right)\right)\)\(+10\text{sin}\left(2{\text{cos}}^{−1}\left(\frac{1}{\sqrt{2}}\right)\right)\)\(+10\text{sin}\left(2{\text{tan}}^{−1}\left(2\right)\right)\) is
[JEE Advanced 2026]
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\(\text { If } \alpha>\beta>\gamma>0 \text {, then find } \cot ^{-1}\left(\frac{1+\alpha \beta}{\alpha-\beta}\right)+\cot ^{-1}\left(\frac{1+\beta \gamma}{\beta-\gamma}\right)+\cot ^{-1}\left(\frac{1+\gamma \alpha}{\gamma-\alpha}\right)\) (24 Jan, Shift II, Memory Based)
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If \(\alpha\) and \(\beta\) are real numbers such that \(\sec ^2\left(\tan ^{-1}(\alpha)\right)+\operatorname{cosec}^2\left(\cot ^{-1}(\beta)\right)=36\) and \(\alpha+\beta=8\), then \(\left(\alpha^2+\beta\right)\) is \((\alpha>\beta)\) (24 Jan, Shift I, Memory Based)
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Using the principal values of the inverse trigonometric functions the sum of the maximum and the minimum values of \(16\left({\left({\sec }^{-1}x\right)}^{2}+{\left({\csc }^{-1}x\right)}^{2}\right)\) is:
[JEE Main 2025, 22 Jan (Shift 1)]
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Considering the principal values of the inverse trigonometric functions, \({\sin }^{-1}\left(\frac{\sqrt{3}}{2}x+\frac{1}{2}\sqrt{1-{x}^{2}}\right),-\frac{1}{2} [JEE Main 2025, 4 Apr (Shift 1)]
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If the domain of the function \(f(x)=\cos ^{-1}\left(\frac{2-|x|}{4}\right)+\left\{\log _{ e }(3-x)\right\}^{-1}\) is \([-\alpha, \beta)-\{\gamma\}\), then \(\alpha+\beta+\gamma\) is equal to :
[JEE Main 2024, 30 Jan (Shift 1)]
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\(\cos \left({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{33}{65}\right)\) is equal to :
[JEE Main 2025, 28 Jan (Shift 1)]
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\(\text{ If }{\cos }^{-1}x=\pi +{\sin }^{-1}x+{\sin }^{-1}(2x-1)\text{,}\\ \text{then find the sum of all values of ' }x\text{ '. }\)
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Considering the principal values of the inverse trigonometric functions, \({\sin }^{-1}\left(\frac{\sqrt{3}}{2}x+\frac{1}{2}\sqrt{1-{x}^{2}}\right),-\frac{1}{2} [JEE Main 2025, 4 Apr (Shift 1)]
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If \(\frac{\pi }{2}\leq x\leq \frac{3\pi }{4}\) , then \({\cos }^{-1}\left(\frac{12}{13}\cos x+\frac{5}{13}\sin x\right)\) is equal to
[JEE Main 2025, 23 Jan (Shift 1)]
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\(\cos \left({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{33}{65}\right)\) is equal to :
[JEE Main 2025, 28 Jan (Shift 1)]
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The value of \({\cot }^{-1}\left(\frac{\sqrt{1+{\tan }^{2}(2)}-1}{\tan (2)}\right)-{\cot }^{-1}\) \(\left(\frac{\sqrt{1+{\tan }^{2}\left(\frac{1}{2}\right)}+1}{\tan \left(\frac{1}{2}\right)}\right)\) is equal to
[JEE Main 2025, 8 Apr (Shift 1)]
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\(\cos \left({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{33}{65}\right)\text{ is equal to: }\) (28 Jan, Shift I, Memory Based)
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The value of \(\tan \left(2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right)\) is equal to:
[JEE Main 2021, 20 Jul (Shift 2)]
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If \({\cot }^{-1}(\alpha )={\cot }^{-1}2+{\cot }^{-1}8+{\cot }^{-1}18+{\cot }^{-1}32+\ldots .\) upto 100 terms, then \( \alpha \) is :
[JEE Main 2021, 17 Mar (Shift 1)]
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The sum of possible values of \( x \) for \( \tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1}\left(\frac{8}{31}\right) \) is
[JEE Main 2021, 17 Mar (Shift 1)]
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A possible value of \( \tan \left(\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8}\right) \) is:
[JEE Main 2021, 24 Feb (Shift 2)]
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\({\tan }^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+{\sec }^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right)\) is equal to
[JEE Main 2023, 24 Jan (Shift 1)]
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The value of \( \tan \left(2 \tan ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)\right) \) is equal to:
[JEE Main 2021, 20 Jul (Shift 2)]
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If \( S \) is the sum of the first 10 terms of the series \( \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right) \) \( +\tan ^{-1}\left(\frac{1}{21}\right)+\ldots \ldots \) then \( \tan (S) \) is equal to
[JEE Main 2020, 5 Sep (Shift 1)]
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Let S be the set of all solutions of the equation \({\cos }^{-1}(2x)-2{\cos }^{-1}\left(\sqrt{1-{x}^{2}}\right)=\pi ,x\in \left[-\frac{1}{2},\frac{1}{2}\right]\text{. }\) Then \(\sum _{x\in S}2{\sin }^{-1}\left({x}^{2}-1\right)\) is equal to
[JEE Main 2023, 1 Feb (Shift 1)]
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Let \({a}_{1}=1,{a}_{2},{a}_{3},{a}_{4},\ldots .\). be consecutive natural numbers. Then \({\tan }^{-1}\left(\frac{1}{1+{a}_{1}{a}_{2}}\right)+{\tan }^{-1}\left(\frac{1}{1+{a}_{2}{a}_{3}}\right)\) \(+\ldots ..+{\tan }^{-1}\left(\frac{1}{1+{a}_{2021}{a}_{2022}}\right)\) is equal to
[JEE Main 2023, 30 Jan (Shift 2)]
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Let \(S=\left\{x\in R:0
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The value of \(\ \cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right) \) at \(\ x=\frac{1}{5} \) is
[JEE Main 2021]
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The range of \(f(x)=4 \sin ^{-1}\left(\frac{x^2}{x^2+1}\right)\) is
[JEE Main 2023, 13 Apr (Shift 2)]
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If \(y(x)={\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right),x\in \left(\frac{\pi }{2},\pi \right)\), then \(\frac{dy}{dx}\) at \(x=\frac{5\pi }{6}\) is:
[JEE Main 2021, 27 Aug (Shift 2)]
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The number of real roots of the equation \({\tan }^{-1}\sqrt{x\left(x+1\right)}+{\sin }^{-1}\sqrt{{x}^{2}+x+1}=\frac{\pi }{4}\)
[JEE Main 2021, 20 Jul (Shift 1)]
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If \( \frac{\sin ^{-1} x}{a}=\frac{\cos ^{-1} x}{b}=\frac{\tan ^{-1} y}{c} ; 0 [JEE Main 2021, 26 Feb (Shift 1)]
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\(\csc \left[2{\cot }^{-1}(5)+{\cos }^{-1}\left(\frac{4}{5}\right)\right]\) is equal to:
[JEE Main 2021, 25 Feb (Shift 2)]
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If \( \sin ^{-1} \frac{\alpha}{17}+\cos ^{-1} \frac{4}{5}-\tan ^{-1} \frac{77}{36}=0,0<\alpha<13 \),
then \( \sin ^{-1}(\sin \alpha)+\cos ^{-1}(\cos \alpha) \) is equal to
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If \({\left({\sin }^{-1}x\right)}^{2}-{\left({\cos }^{-1}x\right)}^{2}=a,0 [JEE Main 2021]
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If the domain of the function\(f(x)={\sec }^{-1}\left(\frac{2x}{5x+3}\right)\) is \([\alpha ,\beta )\cup (\gamma ,\delta ]\), then \(|3\alpha +10(\beta +\gamma )+21\delta |\) is equal to ___
[JEE Main 2023, 10 Apr (Shift 2)]
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If \({\sin }^{-1}\frac{\alpha }{17}+{\cos }^{-1}\frac{4}{5}-{\tan }^{-1}\frac{77}{36}=0,0<\alpha <13\), then \({\sin }^{-1}(\sin \alpha )+{\cos }^{-1}(\cos \alpha )\) is equal to:
[JEE Main 2023, 31 Jan (Shift 1)]
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If \(\sum _{r=1}^{50}{\tan }^{-1}\frac{1}{2{r}^{2}}=p\), then the value of \(\tan p\) is:
[JEE Main 2021, 26 Aug (Shift 2)]
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Let \({S}_{k}=\sum _{r=1}^{k}{\tan }^{-1}\left(\frac{{6}^{r}}{{2}^{2r+1}+{3}^{2r+1}}\right)\) Then \(\lim _{k\to \infty }{S}_{k}\) is equal to:
[JEE Main 2021, 16 Mar (Shift 1)]
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\({\cos }^{-1}(\cos (-5))+{\sin }^{-1}(\sin (6))-{\tan }^{-1}(\tan (12))\) is equal to: (The inverse trigonometric functions take the principal values)
[JEE Main 2021, 1 Sep (Shift 2)]
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Given that the inverse trigonometric functions take principal values only. Then, the number of real values of \( x \) which satisfy \( \sin ^{-1}\left(\frac{3 x}{5}\right)+\sin ^{-1}\left(\frac{4 x}{5}\right)=\sin ^{-1} x \) is equal to :
[JEE Main 2021, 16 Mar (Shift 2)]
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For any \(y\in \mathrm{ℝ}\), let \({\cot }^{-1}(y)\in (0,\pi )\) and \({\tan }^{-1}(y)\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)\) Then the sum of all the solutions of the equation \({\tan }^{-1}\left(\frac{6y}{9-{y}^{2}}\right)+{\cot }^{-1}\left(\frac{9-{y}^{2}}{6y}\right)=\frac{2\pi }{3}\) for \(0<|y|<3\), is equal to:
[JEE Advanced 2023]
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If \(0 [JEE Main 2021, 26 Feb (Shift 2)]
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\(\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)\) is equal to
[JEE Main 2023]
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The domain of the function \( \operatorname{cosec}^{-1}\left(\frac{1+x}{x}\right) \) is:
[JEE Main 2021, 26 Aug (Shift 2)]
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If \(0 [JEE Main 2021, 26 Feb (Shift 2)]
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\( 2 \pi-\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{16}{65}\right) \) is equal to :
[JEE Main 2020, 3 Sep (Shift 1)]
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Let \(S=\left\{x\in R:0 [JEE Main 2023, 1 Feb (Shift 2)]
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The real function \(f(x)=\frac{{\csc }^{-1}x}{\sqrt{x-[x]}}\), where \([x]\) denotes the greatest integer less than or equal to x, is defined for all x belonging to:
[JEE Main 2021, 18 Mar (Shift 1)]
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Let \( g(x)=f(x)+f(1-x) \) and \( f^{\prime \prime}(x)>0, x \in(0,1) \). If \( g \) is decreasing in the interval \( (0, \alpha) \) and increasing in the interval \( (\alpha, 1) \), then the value of \( \tan ^{-1}(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right) \), is equal to
[JEE Main 2023, 10 Apr (Shift 2)]
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If the domain of the function \(f\left(x\right)={\sec }^{-1}\left(\frac{2x}{5x+3}\right)\) is \([\alpha ,\beta )\cup (\gamma ,\delta ]\), then \(|3\alpha +10(\beta +\gamma )+21\delta |\) is equal to ___
[JEE Main 2023, 10 Apr (Shift 2)]
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Let \(f(x)=\cos \left(2{\tan }^{-1}\sin \left({\cot }^{-1}\sqrt{\frac{1-x}{x}}\right)\right)\), \(0 [JEE Main 2021, 26 Aug (Shift 1)]
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The real function \(f\left(x\right)=\frac{{\csc }^{-1}x}{\sqrt{x-[x]}}\), where \([x]\) denotes the greatest integer less than or equal to \(x\), is defined for all \(x\) belonging to:
[JEE Main 2021, 18 Mar (Shift 1)]
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The sum of possible value of \( x \) for \( \tan ^{-1}(x+1)+ \)
\( \cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1}\left(\frac{8}{31}\right) \) is
[JEE Main 2021, 17 Mar (Shift 1)]
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Let \({S}_{k}=\sum _{r=1}^{k}{\tan }^{-1}\left(\frac{{6}^{r}}{{2}^{2r+1}+{3}^{2r+1}}\right)\). Then \(\lim _{k\to \infty }{S}_{k}\) is equal to:
[JEE Main 2021, 16 Mar (Shift 1)]
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Let \((a,b)\subset (0,2\pi )\) be the largest interval for which\({\sin }^{-1}(\sin \theta )-{\cos }^{-1}(\sin \theta )>0,\theta \in (0,2\pi )\) holds. If \(\alpha {x}^{2}+\beta x+{\sin }^{-1}\left({x}^{2}-6x+10\right)+{\cos }^{-1}\left({x}^{2}-6x+10\right)=0\) and \(\alpha -\beta =b-a\), then \(\alpha\) is equal to:
[JEE Main 2023, 31 Jan (Shift 2)]
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Let \( g(x)=f(x)+f(1-x) \) and \( f^{\prime \prime}(x)>0, x \in(0,1) \). If
\( g \) is decreasing in the interval \( (0, \alpha) \) and increasing
in the interval \( (\alpha, 1) \), then the value of
\( \tan ^{-1}(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right) \), is equal to
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If the solution of the equation
\( \log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1 \),
\( x \in\left(0, \frac{\pi}{2}\right) \), is \( \sin ^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right) \), where \( \alpha \beta \) are
integers, then \( \alpha+\beta \) is equal to:
[JEE Main 2023, 30 Jan (Shift 1)]
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If \(\frac{{(x+1)}^{2}}{{x}^{3}+x}=\frac{A}{x}+\frac{Bx+C}{{x}^{2}+1}\), then \({\sin }^{−1}A+{\tan }^{−1}B+{\sec }^{−1}C=\)
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The number of real roots of the equation \({\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{{x}^{2}+x+1}=\frac{\pi }{4}\) is :
[JEE Main 2021, 20 Jul (Shift 1)]
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If \({\left({\sin }^{-1}x\right)}^{2}-{\left({\cos }^{-1}x\right)}^{2}=a,0 [JEE Main 2021, 27 Aug (Shift 1)]
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The domain of the function \(f(x)=\sin ^{-1}\left(\frac{|x|+5}{x^2+1}\right)\) is \((-\infty,-a] \cup[a, \infty)\). Then \(a\) is equal to :
[JEE Main 2020, 2 Sep (Shift 1)]
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If \({\cot }^{-1}(\alpha )={\cot }^{-1}2+{\cot }^{-1}8+{\cot }^{-1}18+{\cot }^{-1}32+\ldots\). upto 100 terms, then \(\alpha\) is :
[JEE Main 2021, 17 Mar (Shift 1)]
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If (sin–1 x)2 – (cos–1 x)2 = a;0 < x < 1;a \(\neq\)0, then the value of 2x2 – 1 is
[JEE Main 2021, 27 Aug (Shift 1)]
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The number of solutions of the equation \( \sin ^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]=x^{2} \), for \( x \in[-1,1] \), and \( [\mathrm{x}] \) denotes the greatest integer less than or equal to \( x \), is :
[JEE Main 2021, 17 Mar (Shift 2)]
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Let \( a_{1}=1, a_{2}, a_{3}, a_{4}, \ldots \) be consecutive natural
numbers.
Then
\( \tan ^{-1}\left(\frac{1}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{1}{1+a_{2} a_{3}}\right)+\ldots+ \)
\( \tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right) \) is equal to
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