Question

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

Answer 1

$\begin{array}{l}\\ \text{LHS=}\frac{\text{tan}\theta }{{(1+{\text{tan}}^2\theta )}^2}+\frac{\text{cot}\theta }{{(1+{\text{cot}}^2\theta )}^2}\\ \text{=}\frac{\text{tan}\theta }{{\left({\text{sec}}^2\theta \right)}^2}+\frac{\text{cot}\theta }{{\left({\text{cosec}}^2\theta \right)}^2}\\ \text{=}\frac{\text{tan}\theta }{{\text{sec}}^4\theta }+\frac{\text{cot}\theta }{{\text{cosec}}^4\theta }\\ \text{=}\frac{\text{sin}\theta }{\text{cos}\theta }\times {\text{cos}}^4\theta +\frac{\text{cos}\theta }{\text{sin}\theta }\times {\text{sin}}^4\theta \\ \text{=sin}\theta {\text{cos}}^3\theta +\text{cos}\theta {\text{sin}}^3\theta \\ \text{=sin}\theta \text{cos}\theta ({\text{cos}}^2\theta +{\text{sin}}^2\theta )\\ \text{=sin}\theta \text{cos}\theta \\ \text{}\text{=RHS}\end{array}$

Hence, LHS = RHS