Question

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

Answer 1

$$\begin{gathered} \begin{array}{l}\\ \left(\text{i}\right)\text{LHS=}\frac{1-\text{sin}\theta }{\text{1+sin}\theta }\\ \text{=}\frac{(1-\text{sin}\theta )(1-\text{sin}\theta )}{\left(\text{1+sin}\theta \right)(1-\text{sin}\theta )}\left[\begin{array}{l}\text{Dividing the numerator and}\\ \text{denominator by}(1-\text{sin}\theta )\end{array}\right]\\ \text{=}\frac{{(1-\text{sin}\theta )}^2}{(1-{\text{sin}}^2\theta )}\\ \text{=}\frac{1+{\text{sin}}^2\theta -2\text{sin}\theta }{{\text{cos}}^2\theta }\\ \text{=}\frac{1}{{\text{cos}}^2\theta }+\frac{{\text{sin}}^2\theta }{{\text{cos}}^2\theta }-\frac{2\text{sin}\theta }{{\text{cos}}^2\theta }\\ {\text{=sec}}^2\theta +{\text{tan}}^2\theta -2\times \frac{1}{\text{cos}\theta }\times \frac{\text{sin}\theta }{\text{cos}\theta }\\ {\text{=sec}}^2\theta +{\text{tan}}^2\theta -2\text{sec}\theta \text{tan}\theta \\ \text{=}{(\text{sec}\theta -\text{tan}\theta )}^2\\ \text{=RHS}\end{array} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \\ \begin{array}{l}\left(\text{ii}\right)\text{LHS=}\frac{1+\text{cos}\theta }{1-\text{cos}\theta }\\ \text{=}\frac{(1+\text{cos}\theta )(1+\text{cos}\theta )}{(1-\text{cos}\theta )(1+\text{cos}\theta )}\left[\begin{array}{l}\text{Dividing the numerator and}\\ \text{denominator by}(1+\text{cos}\theta )\end{array}\right]\\ \text{=}\frac{{(1+\text{cos}\theta )}^2}{(1-{\text{cos}}^2\theta )}\\ \text{=}\frac{1+{\text{cos}}^2\theta +2\text{cos}\theta }{{\text{sin}}^2\theta }\\ \text{=}\frac{1}{{\text{sin}}^2\theta }+\frac{{\text{cos}}^2\theta }{{\text{sin}}^2\theta }+\frac{2\text{cos}\theta }{{\text{sin}}^2\theta }\\ {\text{=cosec}}^2\theta +{\text{cot}}^2\theta +2\frac{1}{\text{sin}\theta }\times \frac{\text{cos}\theta }{\text{sin}\theta }\\ {\text{=cosec}}^2\theta +{\text{cot}}^2\theta +2{\text{cosec}}^2\theta {\text{cot}}^2\theta \\ \text{=}{(\text{cosec}\theta +\text{cot}\theta )}^2\\ \text{=RHS}\end{array} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$