Question

Let $f:A\to B$ be a function defined by $y=f\left(x\right)$ where f is a bijective function, means f is injective (one-one) as well as surjective (onto), then there exist a unique mapping $g:B\to A$ such that $f\left(x\right)=y$ if and only if $g\left(y\right)=x\forall xϵA,yϵB$ Then function g is said to be inverse of f and vice versa so we write $g=f^{-1}:B\to A[\left\{f\right(x),x\}:\{x,f(x\left)\right\}ϵf^{-1}]$when branch of an inverse function is not given (define) then we consider its principal value branch.

If $-1<x<0$,then ${\tan }^{-1}x$ equals?

• A. $\pi -{\cos }^{-1}\left(\sqrt{1-x^2}\right)$
• B. ${\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$
• C. $-{\cos }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$
• D. $cose{c}^{-1}x$

Answer 1

The correct option is B ${\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$

Let ${\tan }^{-1}x=\alpha$
$\Rightarrow \frac{-\pi }{4}<\alpha <0$
$\therefore \tan \alpha =x$ such that $\frac{-\pi }{4}<\alpha <0$
$\therefore \sin \alpha =\frac{x}{\sqrt{1+x^2}}\Rightarrow \alpha ={\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$
$={\cot }^{-1}\left(\frac{1}{x}\right)$
$={\sec }^{-1}\sqrt{1+x^2}$
$=cose{c}^{-1}\frac{\sqrt{1+x^2}}{x}$
$\alpha ={\tan }^{-1}x={\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$