Question

If $(\sec \theta +\tan \theta )=p$. Prove that $\sin \theta =\frac{p^2-1}{p^2+1}$.

Answer 1

$sec\theta +tan\theta =P$ __ (1)

As $se{c}^2\theta -ta{n}^2\theta =1\Rightarrow (sec\theta +tan\theta )(sec\theta -tan\theta )=1$

$\Rightarrow sec\theta -tan\theta =\frac{1}{P}$ __ (2)

$e{q}^n\left(1\right)+e{q}^n\left(2\right)$ $2sec\theta =P+\frac{1}{P}\Rightarrow \frac{2}{cos\theta }=\frac{P^2-1}{P}$

$\Rightarrow cos\theta =\frac{2P}{P^2+1}$ __ (3) $e{q}^n\left(1\right)-e{q}^n\left(2\right)$ $2tan\theta =\frac{P^2-1}{P}$ __ (4)

$e{q}^n\left(3\right)\times \left(4\right)2\times sin\theta =\frac{2P}{P^2+1}\times \frac{P^2-1}{P}$

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