Question

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

Answer 1

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

Hence, LHS = RHS