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$\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=$
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$2 \operatorname{cosec} x$
$\begin{aligned} & \text { (b) } \frac{\sin x}{1+\cos x}=\frac{(\sin x)(1-\cos x)}{\sin ^2 x}=\operatorname{cosec} x-\cot x \\ & \text { and } \frac{1+\cos x}{\sin x}=\frac{1}{\operatorname{cosec} x-\cot x}=\operatorname{cosec} x+\cot x \\ & \therefore \quad \frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=2 \operatorname{cosec} x\end{aligned}$
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