Prove that \( \frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \text{cosec} \theta \)

To prove\( \frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \text{ cosec} \theta \) Solution: L.H.S  \(= \frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}\) By taking LCM of denominator, we get \(= \frac{\sin^2 \theta+(1+\cos \theta)^2}{(1+\cos \theta) \sin \theta}\) Splitting \((1+\cos \theta)^2\)  by using the identity \((a+b)^2=a^2+b^2+2ab\), we get \(= \frac{\sin^2 \theta+1+\cos^2 \theta +2\cos \theta}{(1+\cos \theta) \sin \theta}\) \(= \frac{\sin^2 \theta+\cos^2 \theta +1+2\cos \theta}{(1+\cos \theta) \sin \theta}\) \(= \frac{1 +1+2\cos \theta}{(1+\cos \theta) \sin \theta}\) \(= \frac{2+2\cos \theta}{(1+\cos \theta) \sin \theta}\) \(= \frac{2(1+\cos \theta)}{(1+\cos \theta) \sin \theta}\) \(= \frac{2}{ \sin \theta}\) \(=2 \text{cosec} \theta \)       [As  \( \text{cosec} \theta =\frac{1}{\sin \theta}\)] = R.H.S Hence,  \( \frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \text{ cosec} \theta \)  proved.