Correct Answer – Option 1 : 1/2

Concept:

Trigonometric Identities:

sin² θ + cos² θ = 1.

sin 2θ = 2 sin θ cos θ.

cos 2θ = cos² θ – sin² θ.

Calculation:

Let us express \(\rm\dfrac{\sin x}{1 + \cos x}\) in terms of tan x.

\(\rm \dfrac{\sin x}{1 + \cos x}=\dfrac{2\sin \tfrac{x}{2}\cos \tfrac{x}{2}}{\left (\cos^2 \tfrac{x}{2}+\sin^2 \tfrac{x}{2} \right ) + \left (\cos^2 \tfrac{x}{2}-\sin^2 \tfrac{x}{2} \right )}\)

= \(\rm \dfrac{2\sin \tfrac{x}{2}\cos \tfrac{x}{2}}{2\cos^2 \tfrac{x}{2}}= \dfrac{\sin \tfrac{x}{2}}{\cos \tfrac{x}{2}}=\tan \dfrac{x}{2}\).

∴ \(\rm f(x)=\tan^{-1} \left[\dfrac{\sin x}{1 + \cos x}\right]= \tan^{-1}\left (\tan \dfrac{x}{2} \right )=\dfrac{x}{2}\).

And, the first derivative of f(x) = f'(x) = \(\rm \dfrac{d}{dx}\left (\dfrac{x}{2} \right )=\dfrac{1}{2}\).