Question

What is the range of the function $y=\tan (\sin x+\cos x),$

Find the value of $\sqrt{2}ta{n}^{-1}R$ where $R$ is the highest value in the range.

Answer 1

Clearly Domain is all real number i.e $R$
Also we know $-\sqrt{2}\le \sin x+\cos x\le \sqrt{2}$

$⟹\tan (-\sqrt{2})\le \tan (\sin x+\cos x)\le \tan \left(\sqrt{2}\right)$

$⟹-\tan \sqrt{2}\le y\le \tan \sqrt{2}$

Hence Range is, $-\tan \sqrt{2}\le y\le \tan \sqrt{2}$

Thus $\sqrt{2}ta{n}^{-1}\left(tan\sqrt{2}\right)$ is $2$