Question

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

Answer 1

$\begin{array}{l}\\ \text{LHS=}\frac{\text{sin}\theta }{\left(\text{1+cos}\theta \right)}+\frac{\left(\text{1+cos}\theta \right)}{\text{sin}\theta }\\ \text{=}\frac{{\text{sin}}^2\theta +{\left(\text{1+cos}\theta \right)}^2}{\left(\text{1+cos}\theta \right)\text{sin}\theta }\\ \text{=}\frac{{\text{sin}}^2\theta +1+{\text{cos}}^2\theta +2\text{cos}\theta }{\left(\text{1+cos}\theta \right)\text{sin}\theta }\\ \text{=}\frac{1+1+2\text{cos}\theta }{\left(\text{1+cos}\theta \right)\text{sin}\theta }\\ \text{=}\frac{2+2\text{cos}\theta }{\left(\text{1+cos}\theta \right)\text{sin}\theta }\\ \text{=}\frac{2\left(\text{1+cos}\theta \right)}{\left(\text{1+cos}\theta \right)\text{sin}\theta }\\ \text{=}\frac{2}{\text{sin}\theta }\\ \text{=2cosec}\theta \\ \text{=RHS}\end{array}$

Hence, LHS = RHS