Question

$y={\tan }^{-1}\frac{\sin x}{1+\cos x}$ then prove that $\frac{dy}{dx}=\frac{1}{2}$

Answer 1

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

$=ta{n}^{-1}\frac{sin\left(\frac{2x}{2}\right)}{2co{s}^2\left(x/2\right)}$

$=ta{n}^{-1}\frac{2sin\left(x/2\right)cos\left(x/2\right)}{2co{s}^{2}x/2}$

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

$=ta{n}^{-1}tan\left(x/2\right)$

$y=\frac{x}{2}$

Differentiate with respect to $x$ on both sides we get

$\frac{dy}{dx}=\frac{1}{2}$

Hence proved.