Prove the following

Question:

$\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \operatorname{cosec} \theta$

Solution:

$\mathrm{LHS}=\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=\frac{\sin ^{2} \theta+(1+\cos \theta)^{2}}{\sin \theta(1+\cos \theta)}$

$=\frac{\sin ^{2} \theta+1+\cos ^{2} \theta+2 \cos \theta}{\sin \theta(1+\cos \theta)} \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=\frac{1+1+2 \cos \theta}{\sin \theta(1+\cos \theta)} \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

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

$=2 \operatorname{cosec} \theta=\mathrm{RHS}$ $\left[\because \operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$

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