Inverse Trigonometric Functions: Class 12 PYQs 2016–2026 with solutions

If $2{\cos }^{-1}x=y$, then

The graph of a trigonometric function is as shown. Which of the following will represent graph of its inverse ?

If $y={\sin }^{-1}x$, $-1\le x\le 0$, then the range of y is

Assertion (A) : The range of the function $f(x)=2{\sin }^{-1}x+\frac{3\pi }{2}$, where $x\in [-1,1]$, is $\left[\frac{\pi }{2},\frac{5\pi }{2}\right]$.

Reason (R) : The range of the principal value branch of ${\sin }^{-1}(x)$ is [0, π].

Assertion (A) : Maximum value of $({\cos }^{-1}x{)}^2$ is ${\pi }^2$.

Reason (R) : Range of the principal value branch of ${\cos }^{-1}x$ is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.

The principal value of ${\tan }^{-1}\left(\tan \frac{3\pi }{5}\right)$ is

${\tan }^{-1}3+{\tan }^{-1}\lambda ={\tan }^{-1}\left(\frac{3+\lambda }{1-3\lambda }\right)$ is valid for what values of λ ?

The range of the principal value branch of the function $y={\sec }^{-1}x$ is ________.

The principal value of ${\cos }^{-1}\left(-\frac{1}{2}\right)$ is ________.

Find the value of ${\tan }^{-1}\sqrt{3}-{\cot }^{-1}(-\sqrt{3})$.

Simplify : ${\tan }^{-1}\left(\frac{\cos 2x-\sin 2x}{\cos 2x+\sin 2x}\right)$, $0<x<\frac{\pi }{4}$.

Evaluate : $\tan \left({\sin }^{-1}1-{\cos }^{-1}\left(-\frac{1}{2}\right)\right)$

Evaluate : ${\tan }^{-1}\left[2\sin \left(2{\cos }^{-1}\frac{\sqrt{3}}{2}\right)\right]$

Find the value of ${\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{1}{\sqrt{3}}\right)+{\tan }^{-1}\left[\sin \left(-\frac{\pi }{2}\right)\right]$.

Find the domain of the function $f(x)={\sin }^{-1}(x^2-4)$. Also, find its range.

Evaluate : ${\sin }^{-1}\left(\sin \frac{3\pi }{4}\right)+{\cos }^{-1}\left(\cos \frac{3\pi }{4}\right)+{\tan }^{-1}(1)$

Evaluate ${\sin }^{-1}\left(\sin \frac{3\pi }{4}\right)+{\cos }^{-1}(\cos \pi )+{\tan }^{-1}(1)$.

Draw the graph of ${\cos }^{-1}x$, where $x\in [-1,0]$. Also, write its range.

Prove that ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=2{\cos }^{-1}x$, $\frac{1}{\sqrt{2}}\le x\le 1$.

Prove that : $3{\sin }^{-1}x={\sin }^{-1}(3x-4x^3)$, $x\in \left[-\frac{1}{2},\frac{1}{2}\right]$

If a function f : X → Y defined as f(x) = y is one-one and onto, then we can define a unique function g : Y → X such that g(y) = x, where x ∈ X and y = f(x), y ∈ Y. Function g is called the inverse of function f. The domain of sine function is R and function sine : R → R is neither one-one nor onto. The following graph shows the sine function.

Let sine function be defined from set A to [– 1, 1] such that inverse of sine function exists, i.e., ${\sin }^{-1}x$ is defined from [– 1, 1] to A.
(i) If A is the interval other than principal value branch, give an example of one such interval. (1)
(ii) If ${\sin }^{-1}(x)$ is defined from [– 1, 1] to its principal value branch, find the value of ${\sin }^{-1}\left(-\frac{1}{2}\right)-{\sin }^{-1}(1)$. (1)
(iii) (a) Draw the graph of ${\sin }^{-1}x$ from [– 1, 1] to its principal value branch. (2)
OR
(iii) (b) Find the domain and range of $f(x)=2{\sin }^{-1}(1-x)$. (2)

Prove that ${\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9}=\frac{1}{2}{\sin }^{-1}\left(\frac{4}{5}\right)$.

Solve : ${\tan }^{-1}4x+{\tan }^{-1}6x=\frac{\pi }{4}$.

Find the value of $\sin \left({\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3}\right)$.

If ${\tan }^{-1}\frac{x-3}{x-4}+{\tan }^{-1}\frac{x+3}{x+4}=\frac{\pi }{4}$, then find the value of x.

Solve for x : ${\tan }^{-1}(x-1)+{\tan }^{-1}x+{\tan }^{-1}(x+1)={\tan }^{-1}3x$.

Prove that ${\tan }^{-1}\left(\frac{6x-8x^3}{1-12x^2}\right)-{\tan }^{-1}\left(\frac{4x}{1-4x^2}\right)={\tan }^{-1}2x$; $|2x|<\frac{1}{\sqrt{3}}$.

Find the domain of $g(x)={\cos }^{-1}(x^2-1)$. Hence, find the value of x for which $g(x)=\frac{\pi }{3}$. Also, write the range of ${\cos }^{-1}x$ other than its principal branch.