Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: Exercise 1

The value of ${\displaystyle \sum _{n=1}^{\infty }}{\cot }^{-1}\left(\frac{\left(n^2+2n\right)\left(n^2+2n+1\right)+1}{2n+2}\right)$ is equal to

If ${\sin }^{-1}a+{\sin }^{-1}b+{\sin }^{-1}c=\pi$, then the value of $a\sqrt{\left(1-a^2\right)}+b\sqrt{\left(1-b^2\right)}+c\sqrt{\left(1-c^2\right)}$ will be

An ordered triplet solution $(x, y, z)$ with $x,y,z\in (0,1)$ and satisfying $x^2+y^2+z^2+2xyz=1$ is -

For the equation $2x=\tan \left(2{\tan }^{-1}a\right)+2\tan \left({\tan }^{-1}a+{\tan }^{-1}{a}^3\right)$, which of the following is/are invalid?

If the range of the function $f\left(x\right)={\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3$ is $\left[a, b\right]$ then the value of $\frac{b}{a}$ is

If the value of series ${\sin }^{-1}\frac{1}{\sqrt{2}}+{\sin }^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+...+{\sin }^{-1}\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n\left(n+1\right)}}+.....\infty$ is a then the value of $\frac{\pi }{a}$ is

Number of positive integral ordered pairs $\left(x,y\right)$ which satisfy the equation ${\tan }^{-1}x+{\cos }^{-1}\frac{y}{\sqrt{1+y^2}}={\sin }^{-1}\frac{3}{\sqrt{10}}$ is.

The value $k$ for which there is a triangle whose angles have measure ${\tan }^{-1}\left(\frac{1}{2}\right),{\tan }^{-1}\left(\frac{1}{2}+k\right)$ and ${\tan }^{-1}\left(\frac{1}{2}+2k\right)$, is