Question

Find the range of

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

• A. $[0,1]$
• B. $[-1,0]$
• C. $[-2,2]$
• D. $\left\{0\right\}$

Answer 1

The correct option is D $\left\{0\right\}$

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

$\Rightarrow f\left(x\right)=\frac{\sin x}{\sqrt{1+{\tan }^{2}x}}-\frac{\cos x.\tan x}{\sqrt{1+{\tan }^{2}x}}=0$

Hence range of $f\left(x\right)$ is $\left\{0\right\}$