Question

$\frac{\mathrm{sin\theta }}{(\mathrm{cot\theta }+\mathrm{cosec\theta })}-\frac{\mathrm{sin\theta }}{(\mathrm{cot\theta }-\mathrm{cosec\theta })}=2$

Answer 1

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