JEE Main Inverse Trigonometric Functions Practice Questions With Solutions

Given that the inverse trigonometric function assumes principal values only. Let $x,y$x , y be any two real numbers in $[-1,1]$\left[\right. - 1 , 1 \left]\right. such that ${\cos }^{-1}x-{\sin }^{-1}y=\alpha ,\frac{-\pi }{2}\le \alpha \le \pi$cos^{- 1} ⁡ x - sin^{- 1} ⁡ y = \alpha , \frac{- \pi}{2} \leq \alpha \leq \pi.Then, the minimum value of $x^2+y^2+2xy\sin \alpha$x^{2} + y^{2} + 2 x y sin ⁡ \alpha is

If the domain of the function ${\sin }^{-1}\left(\frac{3x-22}{2x-19}\right)+{\log }_{\mathrm{e}}\left(\frac{3x^2-8x+5}{x^2-3x-10}\right)$sin^{- 1} ⁡ \left(\right. \frac{3 x - 22}{2 x - 19} \left.\right) + log_{e} ⁡ \left(\right. \frac{3 x^{2} - 8 x + 5}{x^{2} - 3 x - 10} \left.\right) is $(\alpha ,\beta ]$\left(\right. \alpha , \beta \left]\right., then $3\alpha +10\beta$3 \alpha + 10 \beta is equal to:

If $a={\sin }^{-1}(\sin (5))$a = sin^{- 1} ⁡ \left(\right. sin ⁡ \left(\right. 5 \left.\right) \left.\right) and $b={\cos }^{-1}(\cos (5))$b = cos^{- 1} ⁡ \left(\right. cos ⁡ \left(\right. 5 \left.\right) \left.\right), then $a^2+b^2$a^{2} + b^{2} is equal to

For $\alpha ,\beta ,\gamma \ne 0$\alpha , \beta , \gamma \neq 0, if ${\sin }^{-1}\alpha +{\sin }^{-1}\beta +{\sin }^{-1}\gamma =\pi$sin^{- 1} ⁡ \alpha + sin^{- 1} ⁡ \beta + sin^{- 1} ⁡ \gamma = \pi and $(\alpha +\beta +\gamma )(\alpha -\gamma +\beta )=3\alpha \beta$\left(\right. \alpha + \beta + \gamma \left.\right) \left(\right. \alpha - \gamma + \beta \left.\right) = 3 \alpha \beta, then $\gamma$\gamma equals

Let $x=\frac{m}{n}$x = \frac{m}{n} ($m,n$m , n are co-prime natural numbers) be a solution of the equation $\cos (2{\sin }^{-1}x)=\frac{1}{9}$cos ⁡ \left(\right. 2 sin^{- 1} ⁡ x \left.\right) = \frac{1}{9} and let $\alpha ,\beta (\alpha >\beta )$\alpha , \beta \left(\right. \alpha > \beta \left.\right) be the roots of the equation $m{x}^2-nx-m+n=0$m x^{2} - n x - m + n = 0. Then the point $(\alpha ,\beta )$\left(\right. \alpha , \beta \left.\right) lies on the line

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$x satisfying ${\tan }^{-1}(x)+{\tan }^{-1}(2x)=\frac{\pi }{4}$tan^{- 1} ⁡ \left(\right. x \left.\right) + tan^{- 1} ⁡ \left(\right. 2 x \left.\right) = \frac{\pi}{4} is :

Let $S=\left\{x\in R:0<x<1\mathrm{and}2{\tan }^{-1}\left(\frac{1-x}{1+x}\right)={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}$S = \left{\right. x \in R : 0 < x < 1 and 2 tan^{- 1} \left(\right. \frac{1 - x}{1 + x} \left.\right) = cos^{- 1} \left(\right. \frac{1 - x^{2}}{1 + x^{2}} \left.\right) \left.\right}.

If $\mathrm{n}(\mathrm{S})$n \left(\right. S \left.\right) denotes the number of elements in $\mathrm{S}$S then :

Let $S$S be the set of all solutions of the equation ${\cos }^{-1}(2x)-2{\cos }^{-1}\left(\sqrt{1-x^2}\right)=\pi ,x\in [-\frac{1}{2},\frac{1}{2}]$cos^{- 1} ⁡ \left(\right. 2 x \left.\right) - 2 cos^{- 1} ⁡ \left(\right. \sqrt{1 - x^{2}} \left.\right) = \pi , x \in \left[\right. - \frac{1}{2} , \frac{1}{2} \left]\right.. Then $\sum _{x\in S}2{\sin }^{-1}(x^2-1)$\underset{x \in S}{\sum} 2 sin^{- 1} ⁡ \left(\right. x^{2} - 1 \left.\right) is equal to :