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Question
Differentiate \[\tan^{- 1} \left( \frac{\cos x}{1 + \sin x} \right)\] with respect to \[\sec^{- 1} x\] ?
Solution
\[\text { Let, u }= \tan^{- 1} \left( \frac{\cos x}{1 + \sin x} \right)\]
\[ \Rightarrow u = \tan^{- 1} \left[ \tan\left( \frac{\pi}{4} - \frac{x}{2} \right) \right]\]
\[ \Rightarrow u = \frac{\pi}{4} - \frac{x}{2}\]
Differentiating it with respect to x,
\[\frac{du}{dx} = 0 - \left( \frac{1}{2} \right)\]
\[\frac{du}{dx} = - \frac{1}{2} . . . \left( i \right)\]
\[\text { Let, v } = se c^{- 1} x\]
Differentiating it with respect to x,
\[\frac{dv}{dx} = \frac{1}{x\sqrt{x^2 - 1}} . . . \left( ii \right)\]
\[\text { Dividing equation } \left( i \right) \text { by}\left( ii \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = - \frac{1}{2} \times \frac{x\sqrt{x^2 - 1}}{1}\]
\[\frac{du}{dv} = \frac{- x\sqrt{x^2 - 1}}{2}\]
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