Question

# ( frac{sin theta}{1+cos theta}+frac{1+cos theta}{operatorname{lin} theta}=operatorname{cosec} theta )

Solution

( begin{aligned} L H S &=frac{sin theta}{1+cos theta}+frac{1+cos theta}{sin theta} &=frac{sin theta(1-cos theta)}{(1+cos theta)(1-cos theta)}+frac{1+cos theta}{sin theta} &=frac{sin theta(1-cos theta)+frac{1+cos theta}{sin theta}}{1-cos ^{2} theta} &=frac{sin theta(1-cos theta)}{sin ^{2} theta}+frac{1+cos theta}{sin theta} &=frac{1-cos theta}{sin theta}+frac{1+cos theta}{sin theta}=frac{2}{sin theta} &=2 operatorname{cosec} theta=end{aligned} )