Question

If $y={\tan }^{-1}\frac{\cos x}{1+\sin x}$ then $\frac{dy}{dx}=$

• A. $1$
• B. $-1$
• C. $\frac{-1}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{-1}{2}$

Given,

$y=ta{n}^{-1}\frac{cosx}{1+sinx}$

$\frac{dy}{dx}=\frac{d}{dx}\left(ta{n}^{-1}\frac{cosx}{1+sinx}\right)$

$f={\tan }^{-1}\left(u\right),u=\frac{\cos \left(x\right)}{1+\sin \left(x\right)}$

$=\frac{d}{du}\left({\tan }^{-1}\left(u\right)\right)\frac{d}{dx}\left(\frac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)$

$=\frac{1}{u^2+1}\cdot \frac{1}{-1-\sin \left(x\right)}$

$=\frac{1}{{\left(\frac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)}^2+1}\cdot \frac{1}{-1-\sin \left(x\right)}$

upon further solving, we get,

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

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

$=-\frac{\sin \left(x\right)+1}{1+1+2\sin x}$

$=-\frac{\sin \left(x\right)+1}{2+2\sin x}$

$=-\frac{\sin \left(x\right)+1}{2(1+\sin x)}$

$=-\frac{1}{2}$