Question

\[\frac{d}{dx} \left\{ \tan^{- 1} \left( \frac{\cos x}{1 + \sin x} \right) \right\} \text { equals }\] ______________ .

Options

• 1/2

• -1/2

• 1

• -1

Answer 1

− 1/2

\[\text { Let u }= \tan^{- 1} \left( \frac{\cos x}{1 + \sin x} \right)\]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2}} \right)\]
\[ \Rightarrow u = \tan^{- 1} \frac{\left( \cos\frac{x}{2} - \sin\frac{x}{2} \right)\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right)}{\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right)^2}\]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}} \right)\]
\[ \Rightarrow u = \tan^{- 1} \left[ \frac{1 - \tan\frac{x}{2}}{1 + \tan\frac{x}{2}} \right]\]
\[ \Rightarrow u = \tan^{- 1} \left[ \frac{\tan\frac{\pi}{4} - \tan\frac{x}{2}}{1 + \tan\frac{\pi}{4} \times \tan\frac{x}{2}} \right]\]
\[ \Rightarrow u = \tan^{- 1} \left[ \tan\left( \frac{\pi}{4} - \frac{x}{2} \right) \right]\]
\[ \Rightarrow u = \frac{\pi}{4} - \frac{x}{2}\]

\[\Rightarrow \frac{du}{dx} = 0 - \left( \frac{1}{2} \right)\]
\[ \Rightarrow \frac{du}{dx} = - \frac{1}{2}\]