Question

If $f\left(x\right)$ = $\frac{sinx}{\sqrt{1+ta{n}^{2}x}}$ - $\frac{cosx}{\sqrt{1+co{t}^{2}x}}$, then find the range of $f\left(x\right)$.

Answer 1

$f\left(x\right)=\frac{\sin x}{\sqrt{1+{\tan }^{2}x}}-\frac{\cos x}{\sqrt{1+{\cot }^{2}x}}$

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

$=\frac{\sin x}{\sqrt{\frac{1}{{\cos }^{2}x}}}-\frac{\cos x}{\sqrt{\frac{1}{{\sin }^{2}x}}}$

$=\sin x\cos x-\cos x\sin x=0$

Range of $f\left(x\right)$ is only $0$.