Question

$(1+{\tan }^2\mathrm{\theta })(1+{\cot }^2\mathrm{\theta })=\frac{1}{({\sin }^2\mathrm{\theta }-{\sin }^4\mathrm{\theta })}$

Answer 1

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

Hence, LHS = RHS