#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.2 question 25 maths textbook solution

Answer: $\operatorname{cosec} x$

Hint: You must know the rules of solving derivative of logarithm trigonometric function.

Given: $\log \left(\frac{\sin x}{1+\cos x}\right)$

Solution:

Let    $y=\log \left(\frac{\sin x}{1+\cos x}\right)$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d}{d x} \log \left(\frac{\sin x}{1+\cos x}\right)$

$\frac{d y}{d x}=\frac{1}{\left(\frac{\sin x}{1+\cos x}\right)} \cdot\left[\frac{(1+\cos x) \frac{d}{d x} \sin x-\sin x \frac{d}{d x}(1+\cos x)}{(1+\cos x)^{2}}\right]$ $\cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

$\frac{d y}{d x}=\frac{1+\cos x}{\sin x} \cdot\left[\frac{(1+\cos x)(\cos x)-\sin x(-\sin x)}{(1+\cos x)^{2}}\right]$

$\frac{d y}{d x}=\left(\frac{1+\cos x}{\sin x}\right) \cdot\left[\frac{\cos x+\cos ^{2} x+\sin ^{2} x}{(1+\cos x)^{2}}\right]$

$\frac{d y}{d x}=\left(\frac{1+\cos x}{\sin x}\right)\left(\frac{\cos x+1}{(1+\cos x)^{2}}\right)$

\begin{aligned} &\frac{d y}{d x}=\frac{(1+\cos x)^{2}}{\sin x(1+\cos x)^{2}} \\ &\frac{d y}{d x}=\frac{1}{\sin x} \end{aligned}

$\frac{d y}{d x}=\operatorname{cosec} x$

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