Question Number 211321 by efronzo1 last updated on 06/Sep/24
$$\:\:\:\:\underbrace{\:} \\ $$
Answered by som(math1967) last updated on 06/Sep/24
$$\:\frac{\mathrm{1}+{sin}\theta}{{cos}\theta}={p} \\ $$$$\Rightarrow\frac{\left(\mathrm{1}+{sin}\theta\right)^{\mathrm{2}} }{{cos}^{\mathrm{2}} \theta}={p}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\left(\mathrm{1}+{sin}\theta\right)^{\mathrm{2}} }{\left(\mathrm{1}+{sin}\theta\right)\left(\mathrm{1}−{sin}\theta\right)}={p}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}+{sin}\theta}{\mathrm{1}−{sin}\theta}={p}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cosec}\:\theta}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{cosec}\:\theta}}={p}^{\mathrm{2}} \\ $$$$\therefore\:\frac{\mathrm{cosec}\:\theta+\mathrm{1}}{\mathrm{cosec}\:\theta−\mathrm{1}}={p}^{\mathrm{2}} \\ $$