$\left(\frac{1-\cos x}{1+\cos x}\right)$



To find: Differentiation of $\left(\frac{1-\cos x}{1+\cos x}\right)$

Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d \cos x}{d x}=-\sin x$

Let us take $u=(1-\cos x)$ and $v=(1+\cos x)$

$u^{\prime}=\frac{d u}{d x}=\frac{d(1-\cos x)}{d x}=\sin x$

$v^{\prime}=\frac{d v}{d x}=\frac{d(1+\cos x)}{d x}=-\sin x$

Putting the above obtained values in the formula:-

$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

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

$=\frac{\sin x+\sin x \cos x+\sin x-\sin x \cos x}{(1+\cos x)^{2}}$

$=\frac{2 \sin x}{(1+\cos x)^{2}}$

Ans $)=\frac{2 \sin x}{(1+\cos x)^{2}}$


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