Question

$\frac{\mathrm{tan\theta }}{(1-\mathrm{cot\theta })}+\frac{\mathrm{cot\theta }}{(1-\mathrm{tan\theta })}=(1+\mathrm{sec\theta }\mathrm{cosec\theta })$

Answer 1

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