Question

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

Answer 1

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