Question

lf $y=\frac{{\sin }^{2}x}{1+cotx}+\frac{{\cos }^{2}x}{1+tanx}$ then $y^{\prime }\left(x\right)$ is equal to

• A. $\cos 2x$
• B. $-\cos 2x$
• C. $2{\cos }^{2}x$
• D. ${\cos }^{3}x$

Answer 1

Answer: (B)

$-\cos 2x$

Given, $y=\frac{{\sin }^{2}x}{1+cotx}+\frac{{\cos }^{2}x}{1+tanx}$
$=\frac{{\sin }^{3}x}{\sin x+\cos x}+\frac{{\cos }^{3}x}{\sin x+\cos x}$
$=\frac{(\sin x+\cos x)({\sin }^{2}x+{\cos }^{2}x-\sin x\cos x)}{\cos x+\sin x}$
$y\left(x\right)=1-\frac{\sin 2x}{2}$
$\Rightarrow y^{\prime }\left(x\right)=-\cos 2x$
Hence, option 'B' is correct.