Question Number 19659 by thukada last updated on 14/Aug/17
$$\frac{\mathrm{1}+{sec}\theta}{{sec}\theta}=\frac{{sin}^{\mathrm{2}\:} \theta}{\mathrm{1}−{cos}\theta} \\ $$$$ \\ $$
Answered by prakash jain last updated on 14/Aug/17
$$\frac{\mathrm{1}+\mathrm{sec}\:\theta}{\mathrm{sec}\:\theta}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sec}\:\theta} \\ $$$$=\mathrm{1}+\mathrm{cos}\:\theta=\frac{\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\left(\mathrm{1}−\mathrm{cos}\:\theta\right)} \\ $$$$=\frac{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta}{\mathrm{1}−\mathrm{cos}\:\theta}=\frac{\mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{1}−\mathrm{cos}\:\theta}\blacksquare \\ $$