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# Solution for Differentiate Tan − 1 ( Cos X 1 + Sin X ) with Respect to Sec − 1 X ? - CBSE (Commerce) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### 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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Solution Differentiate Tan − 1 ( Cos X 1 + Sin X ) with Respect to Sec − 1 X ? Concept: Simple Problems on Applications of Derivatives.
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