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Question
If \[- \frac{\pi}{2} < x < 0 \text{ and y } = \tan^{- 1} \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}, \text{ find } \frac{dy}{dx}\] ?
Solution
\[\text{ We have, y } = \tan^{- 1} \sqrt{\frac{1 - \cos2x}{1 + \cos2x}}\]
\[ \Rightarrow y = \tan^{- 1} \sqrt{\frac{2 \sin^2 x}{2 \cos^2 x}}\]
\[ \Rightarrow y = \tan^{- 1} \sqrt{\tan^2 x}\]
\[ \Rightarrow y = \tan^{- 1} \left( \tan x \right) .......\left[ \because \tan^{- 1} \left( \tan x \right) = - x , \text{ if }x \in \left( - \frac{\pi}{2}, 0 \right) \right]\]
\[ \Rightarrow y = - x\]
\[\Rightarrow \frac{dy}{dx} = - 1\]
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