Differentiate $\tan^{-1}\!\left(\dfrac{1+\cos x}{\sin x}\right)$ with respect to $x$.

CBSE Class 12 · 2018 · DifferentiationLong Answer

Differentiate

tan^{- 1} (\frac{1 + cos x}{sin x})

with respect to

x

Answer: $- \frac{1}{2}$

Use the half-angle forms

1 + cos x = 2 cos^{2} \frac{x}{2}

and

sin x = 2 sin \frac{x}{2} cos \frac{x}{2}

\frac{1 + cos x}{sin x} = \frac{2 cos ^{2} \frac{x}{2}}{2 sin \frac{x}{2} cos \frac{x}{2}} = cot \frac{x}{2} = tan (\frac{π}{2} - \frac{x}{2})

Hence

y = tan^{- 1} [tan (\frac{π}{2} - \frac{x}{2})] = \frac{π}{2} - \frac{x}{2}

.
Differentiate:

\frac{d y}{d x} = - \frac{1}{2}

CBSE Class 12 Board · 2018 · Subjective