Inverse Trigonometry JEE Advanced Questions: Weightage, PYQs, and Preparation Tips

Inverse Trigonometry JEE Advanced Last 10 Years Most Asked Topics

Inverse Trigonometry JEE Advanced Questions asked in the last 10 years include different topics. The table below is prepared based on JEE Main 2026 exam analysis, which shows the most asked JEE Advanced inverse trigonometry question topics.

JEE Main and JEE Advanced Inverse Trigonometric Functions Weightage

The JEE Main previous year trends and analysis show that the number of questions asked from Inverse trigonometry varies year to year. The table given below is prepared based on JEE Main 2026 exam analysis.

Important Inverse Trigonometry Formulas and Identities

To score good marks in JEE Advanced Maths, it is really important to revise formulas regularly. Below are some of the most important inverse-trigonometric formulas, which usually come up:

1. Range of Inverse Trigonometric functions

• $\sin ^{-1} x \in[-\pi / 2, \pi / 2]$

• $\cos ^{-1} \mathrm{x} \in[0, \pi]$

• $\tan ^{-1} \mathrm{x} \in(-\pi / 2, \pi / 2)$

2. Basic inverse trigonometric function identities

• $\sin ^{-1} x+\cos ^{-1} x=\pi / 2$

• $\tan ^{-1} x+\tan ^{-1}(1 / x)=\pi / 2(x>0)$

• $\tan ^{-1} x+\tan ^{-1}(1 / x)=-\pi / 2(x<0)$

3. $\tan ^{-1} \mathrm{a}+\tan ^{-1} \mathrm{~b}=\tan ^{-1}[(\mathrm{a}+\mathrm{b}) /(1-\mathrm{ab})]$

4. Values of inverse Trigonometric functions

Also Check: JEE 2027 Important Formulas

Important Topics from Inverse Trigonometry

Inverse Trigonometric function is the most asked chapter in JEE Advanced. Most questions from this chapter are asked from the topics given below:

1. Domain and Range

2. Principal Value

3. Inverse trigonometry functions graphs

4. Inverse trigonometric functions identities

5. Identities, their sum and differences

6. Applications of inverse trigonometric functions

7. Logarithmic transformations

Also Check: JEE Main 2027 Study Guide Complete Notes, Important Concepts, Formulae and Practice Questions

Inverse Trigonometry JEE Advanced Previous Year Questions

JEE Advanced inverse trigonometry questions are the best way to prepare for JEE Advanced 2027. The questions below will help you understand the exam pattern, the key topics, and also the general level of difficulty.

Question 1: 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).

(1) 1

(2) 2

(3) 3

(4) 5

Solution: Option (3)

Given , $\theta=\tan ^{-1}(2 \tan \theta)-\frac{1}{2} \sin ^{-1}\left(\frac{6 \tan \theta}{9+\tan ^2 \theta}\right)$

Let $x=\tan \theta$.

The equation becomes $\theta=\tan ^{-1}(2 x)-\frac{1}{2} \sin ^{-1}\left(\frac{6 x}{9+x^2}\right)$.

Let us first find the value of $\frac{6 x}{9+x^2}$

$
\frac{6 x}{9+x^2}=\frac{2(3 x)}{9+x^2}=\frac{2\left(\frac{x}{3}\right)}{1+\left(\frac{x}{3}\right)^2}
$
Let $y=\frac{x}{3}$.
Then $\frac{6 x}{9+x^2}=\frac{2 y}{1+y^2}=\sin \left(2 \tan ^{-1}(y)\right)=\sin \left(2 \tan ^{-1}\left(\frac{x}{3}\right)\right)$ [$\sin (2 \theta)=\frac{2 \tan (\theta)}{1+\tan ^2(\theta)}$]

The equation becomes $\theta=\tan ^{-1}(2 x)-\frac{1}{2} \sin ^{-1}\left(\sin \left(2 \tan ^{-1}\left(\frac{x}{3}\right)\right)\right)$.

$
\theta=\tan ^{-1}(2 x)-\tan ^{-1}\left(\frac{x}{3}\right)
$

Apply the tangent function to both sides

$
\begin{aligned}
& \tan (\theta)=\tan \left(\tan ^{-1}(2 x)-\tan ^{-1}\left(\frac{x}{3}\right)\right) . \\
& \tan (\theta)=\frac{2 x-\frac{x}{3}}{1+2 x \frac{x}{3}}=\frac{\frac{5 x}{3}}{1+\frac{2 x^2}{3}}=\frac{5 x}{3+2 x^2} .
\end{aligned}
$
Since $x=\tan (\theta)$, we have $x=\frac{5 x}{3+2 x^2}$.

Solve for $x$

$
\begin{aligned}
& x\left(3+2 x^2\right)=5 x . \\
& 3 x+2 x^3=5 x . \\
& 2 x^3-2 x=0 . \\
& 2 x\left(x^2-1\right)=0 . \\
& 2 x(x-1)(x+1)=0 . \\
& x=0,1,-1 .
\end{aligned}
$

Find the values of $\theta$

$
\begin{aligned}
& \tan (\theta)=0 \Longrightarrow \theta=0 \\
& \tan (\theta)=1 \Longrightarrow \theta=\frac{\pi}{4} \\
& \tan (\theta)=-1 \Longrightarrow \theta=-\frac{\pi}{4}
\end{aligned}
$

Therefore, the total number of real solutions is: 3.

Question 2: $

\mathrm{f}_1(\mathrm{x})=\cot ^{-1}\left(\tan \left(\sin ^2 \mathrm{x}\right)\right), \mathrm{f}_2(\mathrm{x})=\tan ^{-1}\left(\cot \left(\cos ^2 \mathrm{x}\right)\right), \mathrm{f}_3(\mathrm{x})=\cot ^{-1}\left(\operatorname { t a n } \left(\cos ^2 \mathrm{x}\right.\right.

$

and $f_4(x)=\tan ^{-1}\left(\cot \left(\sin ^2 x\right)\right)$, then correct option are

(1)$\mathrm{f}_{1}(\mathrm{x})+\mathrm{f}_{2}(\mathrm{x})+\mathrm{f}_{3}(\mathrm{x})+\mathrm{f}_{4}(\mathrm{x})=2(\pi-1)$

(2)$\mathrm{f}_{1}(\mathrm{x})-\mathrm{f}_{2}(\mathrm{x})+\mathrm{f}_{3}(\mathrm{x})-\mathrm{f}_{4}(\mathrm{x})=0$

(3)$\mathrm{f}_{1}(\mathrm{x})-\mathrm{f}_{2}(\mathrm{x})+\mathrm{f}_{3}(\mathrm{x})-\mathrm{f}_{4}(\mathrm{x})=2 \pi$

(4) $\mathrm{f}_{4}(\mathrm{x})>\mathrm{f}_{3}(\mathrm{x}) \Rightarrow \cos 2 \mathrm{x}>0$

Solution: Option (1), (2), (4)

$\mathrm{f}_{1}(\mathrm{x})=\frac{\pi}{2}-\sin ^{2} \mathrm{x}, \mathrm{f}_{2}(\mathrm{x})=\frac{\pi}{2}-\cos ^{2} \mathrm{x}\\\\ \mathrm{f}_{3}(\mathrm{x})-\frac{\pi}{2}-\cos ^{2} \mathrm{x}, \mathrm{f}_{4}(\mathrm{x})=\frac{\pi}{2}-\sin ^{2} \mathrm{x}$

Hence, the answer is Option (1,2,4)

Question 3: Let $\mathrm{f}(\mathrm{x})=\operatorname{cosec}^{-1}\left(\left[\cos ^2 \mathrm{x}+2 \sin ^2 \mathrm{x}\right]\right)$. Then set of points, where $f(x)$ is not continuous is (where [.] denotes greatest integer function)

(1) $\left\{(2 n-1) \frac{\pi}{2}, n \in z\right\}$

(2) $\left\{\frac{\mathrm{n} \pi}{2}, \mathrm{n} \in \mathrm{z}\right\}$

(3) $\left\{(2 n+1) \frac{\pi}{2}, n \in z\right\}$

(4) none of these

Solution: Option (1)

$
f(x)=\operatorname{cosec}-1\left(\left[\cos ^2 x+2 \sin ^2 x\right]\right)=\operatorname{cosec}^{-1}\left(\left[1+\sin ^2 x\right]\right)
$

when

$
\begin{aligned}
& \sin x= \pm 1, f(x)=\operatorname{cosec} \\
& -1(2)=\frac{\pi}{6} \\
& \sin x \neq \pm 1, f(x)=\operatorname{cosec}^{-1}(1)=\operatorname{cosec} \\
& \text { when }
\end{aligned}{ }^{-1}(1)=\frac{\pi}{2} .
$

Hence $f(x)$ is not continuous at odd multiple of $\frac{\pi}{2}$

Question 4: Find the value of $\left.2 \tan ^{-1}(\operatorname{cosec} \theta)+\tan ^{-1}\left(2 \sin \theta \sec ^2 \theta\right)\right)$ if $\sin \theta$ is the only real root of $\mathrm{x}^3+\mathrm{px}^2+\mathrm{qx}+1=0,(\mathrm{p}<\mathrm{q})$ is

(1) $\pi$

(2) $\frac{\pi}{2}$

(3) $-\frac{\pi}{2}$

(4) $- \pi$

Solution: Option (4)

$\begin{aligned} & f(x)=x^3+p x^2+q x+1 \\ & f(0)=1 \Rightarrow \quad f(0)>0 \\ & f(-1)=p-q \Rightarrow \quad f(-1)<0 \\ & \text { Hence } \sin \theta \in(-1,0) \Rightarrow \theta \in\left(-\frac{\pi}{2}, 0\right) \\ & \quad \text { So, } \quad 2 \tan ^{-1}(\operatorname{cosec} \theta)+\tan ^{-1}\left(2 \sin \theta \sec ^2 \theta\right) \\ & =2 \tan ^{-1}\left(\frac{1}{\sin \theta}\right)+\tan ^{-1}\left(\frac{2 \sin \theta}{1-\sin ^2 \theta}\right) \\ & =2 \tan ^{-1}\left(\frac{1}{\sin \theta}\right)+\tan ^{-1}\left(\tan \left(2 \tan ^{-1}(\sin \theta)\right)\right) \\ & =2\left[\tan ^{-1}\left(\frac{1}{\sin \theta}\right)+\tan ^{-1}(\sin \theta)\right] \\ & =2\left[\frac{-\pi}{2}\right]=-\pi \\ & \text { (as } \sin \theta<0)\end{aligned}$

Question 5: If the equation $\sin ^{-1}(\cos x)+\left|x^2-\lambda\right|=\frac{\pi}{2}$ is inconsistent, then value of $\lambda$ can be

(1) $\frac{-1}{2}$

(2) $\frac{-1}{3}$

(3) $\frac{1}{2}$

(4) $\frac{1}{3}$

Solution: Option (1), (2)

$\begin{aligned} & \sin ^{-1}(\cos x)+\left|x^2-\lambda\right|=\frac{\pi}{2} \\ & \frac{\pi}{2}-\sin ^{-1}(\cos x)=\left|x^2-\lambda\right| \\ & \cos ^{-1}(\cos x)=\left|x^2-\lambda\right|\end{aligned}$

For no point of intersection, $\lambda<0$ and $x^2-\lambda=x$ have no solution

$
x^2-x-\lambda=0
$
$
\mathrm{D}<0
$
$
1+4 \lambda<0 \quad \Rightarrow \quad \lambda<-\frac{1}{4}
$

Question 6: If $f(x)=\sin ^{-1}|\sin x|$ and $g(x)=\left(\sin ^{-1}|\sin x|\right)^2$ when, $x \in[0,2 \pi]$, then area between $f(x)$ and $g(x)$ is

(1) $\frac{4}{3}+\frac{\pi^{2}(\pi-1)}{6} \\$

(2) $\frac{1}{6}+\frac{\pi^{3}}{8} \\$

(3) $\frac{3}{4}+\frac{\pi^{2}(\pi-2)}{6} \\$

(4) $\frac{4}{3}+\frac{\pi^{2}(\pi-3)}{6}$

Solution: Option (4)

$\begin{aligned} & \text { Area }=4\left[\int_0^1\left(x-x^2\right)+\int_1^{\pi / 2}\left(x^2-x\right) d x\right] \\ & =4\left[\frac{x^2}{2}-\left.\frac{x^3}{3}\right|_0 ^1+\frac{x^3}{3}-\left.\frac{x^2}{2}\right|_1 ^{\pi / 2}\right] \\ & =\frac{4}{3}+\frac{\pi^2(\pi-3)}{6}\end{aligned}$

Question 7: The area enclosed between $f(x)$ and $\{\mathrm{x}\}$-axis, when $f(x)=\min \left\{\cos ^{-1}(\cos x), \cot ^{-1}(\cot x)\right\}$ and $x \in(\pi, 2 \pi)$ is

(1) $\frac{\pi^{2}}{2} \\$

(2) $\pi^{2} \\$

(3) $\frac{\pi^{2}}{4} \\$

(4) $\frac{\pi^{2}}{3}$

Solution: Option (3)

$Area =\frac{1}{2} \times \pi \times \frac{\pi}{2}=\frac{\pi^{2}}{4}$

Question 8: Which of the following is a rational number

(1) $\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)$

(2) $\log _{2}\left(\sin \left(\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8}\right)\right)$

(3) $\cos \left(\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right)\right)$

(4) $\cos \left(\cot ^{-1} \frac{1}{3}+\cot ^{-1} 3\right)$

Solution: Option (2), (3), (4)

$
\begin{aligned}
& \cos ^{-1} \frac{\sqrt{5}}{3}=\theta \Rightarrow \cos \theta=\frac{\sqrt{5}}{3} \Rightarrow \tan \theta=\frac{2}{\sqrt{5}} \\
& \tan \theta=\frac{2 \tan \frac{\theta}{2}}{1-\tan ^2 \frac{\theta}{2}} \Rightarrow \tan \frac{\theta}{2}=\frac{3-\sqrt{5}}{2}
\end{aligned}
$

B $
\sin ^{-1} \frac{\sqrt{63}}{8}=\theta \Rightarrow \sin \theta=\frac{\sqrt{63}}{8}, \cos \theta=\frac{1}{8}
$

As $\cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}}=\frac{3}{4}$
And $\sin \frac{\theta}{4}=\sqrt{\frac{1-\cos \frac{\theta}{2}}{2}}=\frac{1}{2 \sqrt{2}}$
Now, $\log _2 \sin \left(\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8}\right)=\log _2 \frac{1}{2 \sqrt{2}}=-\frac{3}{2}$

C $\cos \left(\frac{\pi}{2}-\sin ^{-1} \frac{4}{5}\right)=\cos \left(\cos ^{-1} \frac{4}{5}\right)=\frac{4}{5}$

D $\cos \left(\cot ^{-1} \frac{1}{3}+\cot ^{-1} 3\right)=\cos \left(\tan ^{-1} 3+\cot ^{-1} 3\right)=\cos \left(\frac{\pi}{2}\right)=0$

Question 9: Match the column

(1) $p-1, q-4, r-4, s-4$

(2) $p –1, q –4, r-3, s-2$

(3) $p –1, q –3, r-4, s-2$

(4) $p –1, q –2, r-3, s-4$

Solution: Option (1)

$
\begin{aligned}
& x^2-x-\pi+2<0 \\
& x=\frac{1 \pm \sqrt{4 \pi-7}}{2} \\
& \therefore \frac{1-\sqrt{4 \pi-7}}{2}<\mathrm{x}<\frac{1+\sqrt{4 \pi-7}}{2} \quad \therefore \lambda=1 \\
& 2[\mathrm{x}]=\mathrm{x}+2\{\mathrm{x}\}
\end{aligned}
$

(i) If x is an integer, then the equation becomes $2 x=x+0$
i.e. $x=0$ is a solution
(ii) If $x \neq 1$ the equation becomes

$
\begin{aligned}
& 2[\mathrm{x}]=[\mathrm{x}]+\{\mathrm{x}\}+2\{\mathrm{x}\} \text { i.e. } \quad\{\mathrm{x}\}=\frac{1}{3}[\mathrm{x}] \\
& 0<\frac{[\mathrm{x}]}{3}<1 \Rightarrow 0<[\mathrm{x}]<3
\end{aligned}
$

Possible values of $[x]$ are 1,2
If $[x]=1$, then $\{x\}=\frac{1}{3} \quad \therefore \quad x=1+\frac{1}{3}=\frac{4}{3}$
If $[\mathrm{x}]=2$, then $\{\mathrm{x}\}=\frac{2}{3} \quad \therefore \quad \mathrm{x}=\frac{8}{3}$
There are 3 solutions

$
\begin{gathered}
\text { Let } x=\cos \theta, y=\sin \theta \\
\therefore x+y=\cos \theta+\sin \theta \\
\therefore \quad \text { maximum value of } x+y \text { is } \sqrt{2} \\
\mathrm{f}\left(\mathrm{x}+\frac{1}{2}\right)+\mathrm{f}\left(\mathrm{x}-\frac{1}{2}\right)=\mathrm{f}(\mathrm{x}) \\
\mathrm{f}(\mathrm{x}+1)+\mathrm{f}(\mathrm{x})=\mathrm{f}\left(\mathrm{x}+\frac{1}{2}\right) \Rightarrow \mathrm{f}(\mathrm{x}+1)+\mathrm{f}\left(\mathrm{x}-\frac{1}{2}\right)=0 \\
\mathrm{f}\left(\mathrm{x}+\frac{3}{2}\right)=-\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{f}(\mathrm{x}+3)=-\mathrm{f}\left(\mathrm{x}+\frac{3}{2}\right)=\mathrm{f}(\mathrm{x})
\end{gathered}
$

$f(x)$ is periodic with period 3

Question 10: Match the column

(1) $A-r, B-q, C-s, D-t$

(2) $A –p, B –q, C-t, D-s$

(3) $A –p, B –q, C-s, D-t$

(4) $A –r, B-p, C-t, D-s$

Solution: Option (2)

A. The difference $=2-(-2)=4$
B. Let $\mathrm{f}(\mathrm{x})=\mathrm{x}^2-4 \mathrm{x}+3$

$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-4=0 \quad \Rightarrow \mathrm{x}=2 \\
& \mathrm{f}(1)=0, \mathrm{f}(2)=-1, \mathrm{f}(3)=0
\end{aligned}$

$\therefore \mid$ greatest value-least value $\mid=1$
C.

$\tan ^{-1} \frac{1-x}{1+x}=\tan ^{-1} 1-\tan ^{-1} x$

$\therefore \text { greatest value }=\frac{\pi}{4}$

D.

$\therefore \text { greatest value }=\frac{\pi}{2}, \text { least value }=\frac{\pi}{3}$

Question 11. If sum of the series $\cot ^{-1}\left(2 \cdot 1^2\right)+\cot ^{-1}\left(2 \cdot 2^2\right)+\cot ^{-1}\left(2 \cdot 3^2\right) \ldots \ldots \infty$ is equal to $k \pi$ then find the value of $[k]$, while $[\cdot]$ denotes GIF

(1) 0

(2) 1

(3) 2

(4) 3

Solution: Option (1)

$\begin{aligned} \mathrm{s} & =\sum_{\mathrm{r}=1}^{\infty} \cot ^{-1}\left(2 \cdot \mathrm{r}^2\right)=\sum_{\mathrm{r}=1}^{\infty} \tan ^{-1} \frac{1}{2 \mathrm{r}^2} \\ & =\sum_{\mathrm{r}=0}^{\infty} \tan ^{-1}\left(\frac{(2 \mathrm{r}+1)-(2 \mathrm{r}-1)}{1+(2 \mathrm{r}+1)(2 \mathrm{r}-1)}\right) \\ & =\sum_{\mathrm{r}=1}^{\infty} \tan ^{-1}(2 \mathrm{r}+1)-\tan ^{-1}(2 \mathrm{r}-1) \\ & =\left(\tan ^{-1} 3-\tan ^{-1} 1\right)+\left(\tan ^{-1} 5-\tan ^{-1} 3\right)+\left(\tan ^{-1} 7-\tan ^{-1} 5\right)+\ldots \ldots \tan ^{-1} \infty \\ & =-\tan ^{-1} 1+\tan ^{-1} \infty \\ & =-\frac{\pi}{4}+\frac{\pi}{2}=\frac{\pi}{4} \Rightarrow \mathrm{k}=\frac{1}{4}\end{aligned}$

Question 12. $\begin{aligned} & \cot ^{-1}\left(\frac{5+3 \cos 2 \mathrm{x}}{3 \sin 2 \mathrm{x}}\right)+\cot ^{-1}\left(\frac{4}{\tan \mathrm{x}}\right)=\lambda \mathrm{x} \text {, then find the value of } \lambda \text { where } \\ & x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\end{aligned}$

(1) 1

(2) 2

(3) 3

(4) 4

Solution: Option (1)

$\begin{aligned} & \Rightarrow \cot ^{-1}\left(\frac{5+3 \cos 2 \mathrm{x}}{3 \sin 2 \mathrm{x}}\right)+\cot ^{-1}\left(\frac{4}{\tan \mathrm{x}}\right) \\ & \Rightarrow \tan ^{-1}\left(\frac{3 \sin 2 \mathrm{x}}{5+3 \cos 2 \mathrm{x}}\right)+\tan ^{-1}\left(\frac{\tan \mathrm{x}}{4}\right) \\ & \Rightarrow \tan ^{-1}\left(\frac{6 \tan \mathrm{x}}{8+2 \tan ^2 \mathrm{x}}\right)+\tan ^{-1}\left(\frac{\tan \mathrm{x}}{4}\right) \\ & \Rightarrow \tan ^{-1}\left(\frac{\frac{3 \tan x}{4+\tan ^2 x}+\frac{\tan x}{4}}{1-\frac{3 \tan ^2 x}{16+4 \tan ^2 x}}\right) \\ & \Rightarrow \tan ^{-1}(\tan x) \\ & =x\end{aligned}$

Question 13. $\mathrm{S}=\cot ^{-1}\left(\frac{1}{8}\right)+\cot ^{-1}\left(\frac{1}{2}\right)+\cot ^{-1}\left(\frac{9}{8}\right)+\cot ^{-1}(2)+\cot ^{-1}\left(\frac{25}{8}\right)+\ldots$

$ \text { If } \lim _{n \rightarrow \infty} \mathrm{S}=\frac{\mathrm{a}}{\mathrm{b}} \pi$

then the correct option is/are

(1) $a+b=9$

(2) $a-b=5$

(3) $a-b=3$

(4) $a+b=11$

Solution: Options (3) and (4)

$\begin{aligned} & \mathrm{T}_{\mathrm{n}}=\cot ^{-1}\left(\frac{\mathrm{n}^2}{8}\right) \\ & =\tan ^{-1}\left(\frac{8}{\mathrm{n}^2}\right)=\tan ^{-1}\left(\frac{2}{1+\frac{\mathrm{n}^2}{4}-1}\right) \\ & =\tan ^{-1}\left(\frac{\mathrm{n}}{2}+1\right)-\tan ^{-1}\left(\frac{\mathrm{n}}{2}-1\right) \\ & \sum_{\mathrm{n}=1}^{\infty} \mathrm{T}_{\mathrm{n}}=\frac{7 \pi}{4}\end{aligned}$

Question 14: 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:

(1) 6

(2) 7

(3) 8

(4) 9

Solution: Option (2)

$\begin{aligned} & \beta=\frac{1}{3 \alpha}, \tan ^{-1}(1-\alpha)+\tan ^{-1}(1-\beta)=\frac{\pi}{4} \\ & \Rightarrow \tan \left(\tan ^{-1}(1-\alpha)+\tan ^{-1}(1-\beta)\right)=1 \\ & \Rightarrow \frac{1-\alpha+1-\beta}{1-(1-\alpha)(1-\beta)}=1 \\ & \Rightarrow 2-\alpha-\beta=1-(1-\alpha-\beta+\alpha \beta) \\ & \Rightarrow 2-\alpha-\beta=\alpha+\beta-\alpha \beta \\ & \Rightarrow 2(\alpha+\beta)=2+\alpha \beta \\ & \Rightarrow 2(\alpha+\beta)=2+\frac{1}{3}=\frac{7}{3} \\ & \alpha+\beta=\frac{7}{6} \\ & 6(\alpha+\beta)=7\end{aligned}$

Question 15: If $\sin \left(\tan ^{-1}(x \sqrt{2})\right)=\cot \left(\sin ^{-1} \sqrt{1-x^2}\right), x \in(0,1)$, then the value of $x$ is:

(1) $\frac{1}{2}$

(2) $\frac{1}{3}$

(3) $\frac{2}{3}$

(4) $\frac{5}{8}$

Solution: Option (1)

$\begin{aligned} & \frac{x \sqrt{2}}{\sqrt{2 x^2+1}}=\frac{1}{\sqrt{1-x^2}} \\ & 2\left(1-x^2\right)=\left(2 x^2+1\right) \\ & 2-2 x^2=2 x^2+1 \\ & x^2=\frac{1}{4} \\ & x=\frac{1}{2}\end{aligned}$

Also check: JEE Mains 2027: Maths Set of 5 Sample Paper with Solution

Best Books for Inverse Trigonometric JEE Main Questions

Refer to the books given below for solving JEE Advanced inverse trigonometry questions. While preparing for IIT JEE, it is very important to follow the Best Books for JEE Main 2027 because they help properly cover all the concepts.