Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 2: EXERCISE-2

Value of $k$ for which the point $\left(\alpha , {\sin }^{-1}\alpha \right)\left(\alpha >0\right)$ lies inside the triangle formed by $x+y=k$ with co-ordinate axes is -

Solution set of the inequality ${\sin }^{-1}\left(\sin \frac{2x^2+3}{x^2+1}\right)\le \pi -\frac{5}{2}$ is -

Consider two geometric progressions $a_1,a_2,a_3\mathrm{.......}{a}_n$ & $b_1, b_2, b_3,\mathrm{.....}{b}_n$ with $a_r=\frac{1}{b_r}=2^{r-1}$ and another sequence $t_1,t_2,t_3\mathrm{.......}{t}_n$ such that $t_r={\cot }^{-1}\left(2a_r+b_r\right)$ then $\underset{n\to \infty }{\lim }\sum _{r=1}^n{t}_r$ is -

The sum of the infinite terms of the series - ${\cot }^{-1}\left(1^2+\frac{3}{4}\right)+{\cot }^{-1}\left(2^2+\frac{3}{4}\right)+{\cot }^{-1}\left(3^2+\frac{3}{4}\right)+..........$ is equal to -

If numerical value of $\tan \left\{{\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3}\right\}$ is $\frac{a}{b}$, then -

The value of $\cos \left[\frac{1}{2}{\cos }^{-1}\left(\cos \left(-\frac{14\pi }{5}\right)\right)\right]$ is/are

${\tan }^{-1}\left(\frac{1}{4}\right)+{\tan }^{-1}\left(\frac{2}{9}\right)$ equals to

${\sin }^{-1}\frac{3x}{5}+{\sin }^{-1}\frac{4x}{5}={\sin }^{-1}x$, then roots of the equation are -