Question

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

Answer 1

$\begin{array}{l}\\ \left(\text{i}\right)\text{LHS=}\frac{\text{sec}\theta -1}{\text{sec}\theta \text{+1}}\\ \text{=}\frac{\frac{1}{\text{cos}\theta }-1}{\frac{1}{\text{cos}\theta }+1}\\ \text{=}\frac{\frac{1-\text{cos}\theta }{\text{cos}\theta }}{\frac{1+\text{cos}\theta }{\text{cos}\theta }}\\ \text{=}\frac{1-\text{cos}\theta }{1+\text{cos}\theta }\\ \text{=}\frac{(1-\text{cos}\theta )(1+\text{cos}\theta )}{(1+\text{cos}\theta )(1+\text{cos}\theta )}\left\{\begin{array}{l}\text{Dividing the numerator and}\\ \text{denominator by}(1+\text{cos}\theta )\end{array}\right\}\\ \text{=}\frac{1-{\text{cos}}^2\theta }{{(1+\text{cos}\theta )}^2}\\ \text{=}\frac{{\text{sin}}^2\theta }{{(1+\text{cos}\theta )}^2}\\ \text{=RHS}\\ \left(\text{ii}\right)\text{LHS=}\frac{\text{sec}\theta -\text{tan}\theta }{\text{sec}\theta \text{+tan}\theta }\\ \text{=}\frac{\frac{1}{\text{cos}\theta }-\frac{\text{sin}\theta }{\text{cos}\theta }}{\frac{1}{\text{cos}\theta }+\frac{\text{sin}\theta }{\text{cos}\theta }}\\ \text{=}\frac{\frac{1-\text{sin}\theta }{\text{cos}\theta }}{\frac{\text{1+sin}\theta }{\text{cos}\theta }}\\ \text{=}\frac{1-\text{sin}\theta }{\text{1+sin}\theta }\\ \text{=}\frac{(1-\text{sin}\theta )\left(\text{1+sin}\theta \right)}{\left(\text{1+sin}\theta \right)\left(\text{1+sin}\theta \right)}\text{{}\begin{array}{l}\text{Dividing the numerator and}\\ \text{denominator by}\left(\text{1+sin}\theta \right)\end{array}\}\\ \text{=}\frac{(\text{1}-{\text{sin}}^2\theta )}{{\left(\text{1+sin}\theta \right)}^2}\\ \text{=}\frac{{\text{cos}}^2\theta }{{\left(\text{1+sin}\theta \right)}^2}\\ \text{=RHS}\end{array}$