Question

If $\sec A+\tan A=p,$ show that : $\sin A=\frac{p^2-1}{p^2+1}$

Answer 1

$\sec A+\tan A=p$

$\frac{p^2-1}{p^2+1}$

$\Rightarrow \frac{{\sec }^{2}A+{\tan }^{2}A+2\sec A\tan A-1}{{\sec }^{2}A+{\tan }^{2}A+2\sec A\tan A+1}$

$\frac{2{\tan }^{2}A+2\sec A\tan A}{2{\sec }^{2}A+2\sec A\tan A}$

$\frac{\tan A(\tan A+\sec A)}{\sec A(\sec A+\tan A)}$

$=\sin A$.