Question

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

Answer 1

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