Question

The domain and range of $y=f\left(x\right)=\tan (\sin x+\cos x)$ are

• A. $D\left(f\right)=R,R\left(f\right)=[-\tan \sqrt{2},\tan \sqrt{2}]$
• B. $D\left(f\right)=R,R\left(f\right)=[0,\tan \sqrt{2}]$
• C. $D\left(f\right)=R,R\left(f\right)=(-\tan \sqrt{2},\tan \sqrt{2})$
• D. $D\left(f\right)=R,R\left(f\right)=(0,\tan \sqrt{2})$

Answer 1

The correct option is A $D\left(f\right)=R,R\left(f\right)=[-\tan \sqrt{2},\tan \sqrt{2}]$

$y=f\left(x\right)=\tan (\sin x+\cos x)$
$y=\tan \left[\sqrt{2}\sin (x+\frac{\pi }{4})\right]$
The above step has come by using the expansion of $\sin (x+\frac{\pi }{4})=\sin \frac{\pi }{4}\cos x+\cos \frac{\pi }{4}\sin x=\sqrt{2}(\sin x+\cos x)$
Clearly, $f\left(x\right)$ is defined for all real values of $x$ as domain of sine function is $R$.
Range :
$-1\le \sin (x+\frac{\pi }{4})\le 1$
$\Rightarrow -\sqrt{2}\le \sqrt{2}\sin (x+\frac{\pi }{4})\le \sqrt{2}$
$\Rightarrow -\tan \sqrt{2}\tan \left[\sqrt{2}\sin (x+\frac{\pi }{4})\right]\le \tan \sqrt{2}$
$\Rightarrow -\tan \sqrt{2}\le y\le \tan \sqrt{2}$

Range of $f$ is $[-\tan \sqrt{2},\tan \sqrt{2}]$