Question

If $\sec x+\tan x=p$ then prove that $\sin x=\frac{p^2-1}{p^2+1}$ .

Answer 1

Given: $\sec x+\tan x=p$ .... (1)

$\sec x-\tan x=1/p$ $\because Se{c}^{2}x-ta{n}^{2}x=(Secx-tanx)(secx+tanx)=1$

By adding the above equation we got;

$2\sec x=p+\frac{1}{p}=\frac{p^2+1}{p}$

$\sec x=\frac{p^2+1}{2p}$

$\cos x=\frac{2p}{p^2+1}$

${AC}^2={AB}^2+{BC}^2$

${(p^2+1)}^2={\left(2p\right)}^2+{\left(AB\right)}^2$

$p^4+1+2p^2-4p^2={AB}^2$

By solving;

$AB=p^2-1$

$\therefore \sin x=\frac{p^2-1}{p^2+1}$