Question

Prove that $\frac{(1+\mathrm{sin\theta })}{(1-\mathrm{sin\theta })}$ = (sec θ + tan θ)².

Answer 1

$\frac{(1+\mathrm{sin\theta })}{(1-\mathrm{sin\theta })}$= (sec θ + tan θ)²

$$\begin{gathered} \text{LHS=}\frac{(1+\sin \theta )}{(1-\sin \theta )} \\ \text{Multiplying the numerator and denominator by}(1+\sin \theta ),\mathrm{we}\mathrm{get}: \\ \frac{{(1+\sin \theta )}^2}{1-{\sin }^2\theta } \\ \text{=}\frac{1+2\sin \theta +{\sin }^2\theta }{{\cos }^2\theta }[\because {\sin }^2\theta +{\cos }^2\theta =1] \\ {\text{=sec}}^2\theta +2\times \frac{\sin \theta }{\cos \theta }\times \text{sec}\theta +{\tan }^2\theta \\ {\text{=sec}}^2\theta +2\times \tan \theta \times \text{sec}\theta +{\tan }^2\theta \\ \text{=}{\left(\text{sec}\theta \text{+tan}\theta \right)}^2 \\ \text{=RHS} \\ \text{Hence proved}\text{.} \end{gathered}$$