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#### provide solution for RD Sharma maths class 12 chapter Differentiation exercise  10.2 question 26

Answer:$\operatorname{cosec} x$

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

Given: $\log \sqrt{\frac{1-\cos x}{1+\cos x}}$

Solution:

Let   $y=\log \sqrt{\frac{1-\cos x}{1+\cos x}}$

$y=\frac{1}{2} \log \left(\frac{1-\cos x}{1+\cos x}\right)$                                                    using $\log a^{b}=b \log a$

Differentiate with respect to x

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

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

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

Using quotient rule

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

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

$\frac{d y}{d x}=\frac{\sin x}{(1-\cos x)(1+\cos x)}$

\begin{aligned} &\frac{d y}{d x}=\frac{\sin x}{\left(1-\cos ^{2} x\right)} \\ &\frac{d y}{d x}=\frac{\sin x}{\sin ^{2} x} \end{aligned}\begin{aligned} &\frac{d y}{d x}=\frac{1}{\sin x} \\ &\frac{d y}{d x}=\operatorname{cosec} x \end{aligned}