Question

$\frac{\left(1+{\tan }^2\mathrm{\theta }\right)\mathrm{cot\theta }}{{\mathrm{cosec}}^2\mathrm{\theta }}=\tan \theta$

Answer 1

$\begin{array}{l}\\ \text{LHS=}\frac{(1+{\text{tan}}^2\theta )\text{cot}\theta }{{\text{cosec}}^2\theta }\\ \text{=}\frac{{\text{sec}}^2\theta \text{cot}\theta }{{\text{cosec}}^2\theta }\\ \text{=}\frac{\frac{1}{{\text{cos}}^2\theta }\times \frac{\text{cos}\theta }{\text{sin}\theta }}{\frac{1}{{\text{sin}}^2\theta }}\\ \text{=}\frac{1}{\text{cos}\theta \text{sin}\theta }\times {\text{sin}}^2\theta \\ \text{=}\frac{\text{sin}\theta }{\text{cos}\theta }\\ \text{=tan}\theta \\ \text{=RHS}\end{array}$
Hence, L.H.S. = R.H.S.