Question

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

Answer 1

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

∴ LHS = RHS
Hence proved.