Differentiate ^ -1 ( 1+ ) with respect to ^ -1 x

Question

Differentiate $\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$ with respect to $\sec^{-1}\text{x}$

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Answer

Let, $\text{u}=\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}-\frac{\text{x}}{2}\Big)\Big]$
$\Rightarrow\text{u}=\frac{\pi}{4}-\frac{\text{x}}{2}$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0-\big(\frac{1}{2}\big)$
$\frac{\text{du}}{\text{dx}}=-\frac{1}{2}\ .....(\text{i})$
Let, $\text{v}=\sec^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{\text{x}\sqrt{\text{x}^2-1}}\ .....\text{(ii)}$
Differentiating equation (i) by (ii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=-\frac{1}{2}\times\frac{\text{x}\sqrt{\text{x}^2-1}}{1}$
$\frac{\text{du}}{\text{dv}}=\frac{-\text{x}\sqrt{\text{x}^2-1}}{2}$

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