Question

If $co\sec \theta +\cot \theta =P$, then find $\cos \theta$ in terms of $P$.

Answer 1

We know the trignometric identity $cose{c}^2\theta -co{t}^2\theta =1$ which is written as $(cosec\theta -cot\theta )(cosec\theta +cos\theta )=1$

Given $cosec\theta +cos\theta =P\left(1\right)$ , hence $cosec\theta -cos\theta =\frac{1}{P}\left(2\right)$. By adding (1) and (2), we obtain $2cosec\theta =P+\frac{1}{P}=\frac{P^2+1}{P}$

Hence $sin\theta =\frac{2P}{P^2+1}$ and $si{n}^2\theta =\frac{4P^2}{(P^2+1{)}^2}$

$co{s}^2\theta =1-si{n}^2\theta$

$=1-\frac{4P^2}{(P^2+1{)}^2}=\frac{(P^2-1{)}^2}{(P^2+1{)}^2}$

$cos\theta =\frac{P^2-1}{P^2+1}$