f ( x ) = sin x + cos x , g ( x ) = x 2 - 1 then g ( f ( x ) ) is invertible if

$f\left(x\right)=\sin x+\cos x , g\left(x\right)=x^2 -1$

$g\left[f\right(x\left)\right]=g(\sin x+\cos x)={(\sin x+\cos x)}^2 -1$

$= \sin { }^2 x+{\cos }^2 x+2\sin x.\cos x-1$

$= 1+2\sin x.\cos x-1 [\because {\sin }^2 x+{\cos }^2=1]$

$=\sin 2x$

$[\because 2\sin x.\cos x=\sin 2x]$

A function is invertible if it is one-one and onto

Now

$-1\le \sin x \le 1$

$\Rightarrow -1\le \sin 2x \le 1$

$\Rightarrow \frac{-\pi }{2}\le 2x\le \frac{\pi }{2}$

$\Rightarrow \frac{-\pi }{4}\le x\le \frac{\pi }{4}$