Question

If sec $\theta +\tan \theta =p,$ then find the value of $\text{cosec}\theta$.

Answer 1

Given $\text{sec}\theta +\tan \theta =p$

as we know that

${\text{sec}}^2\theta -{\tan }^2\theta =1⟹(\text{sec}\theta -\tan \theta )(\text{sec}\theta +\tan \theta )=1⟹\text{sec}\theta -\tan \theta =\frac{1}{p}$

So $\text{sec}\theta =\frac{1}{2}(p+\frac{1}{p})=\frac{p^2+1}{2p},\tan \theta =\frac{1}{2}(p-\frac{1}{p})=\frac{p^2-1}{2p}$

$\text{cosec}\theta =\frac{1}{\sin \theta }=\frac{\frac{1}{\cos \theta }}{\frac{\sin \theta }{\cos \theta }}=\frac{\text{sec}\theta }{\tan \theta }=\frac{\frac{p^2+1}{2p}}{\frac{p^2-1}{2p}}=\frac{p^2+1}{p^2-1}$