Question

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

Answer 1

$\frac{m^2-1}{m^2+1}=\frac{(\sec A+\tan A{)}^2-1}{(\sec A+\tan A{)}^2+1}$

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

$=\frac{({\sec }^{2}A-1)+{\tan }^{2}A+2\sec A\tan A}{{\sec }^{2}A+({\tan }^{2}A+1)+2\sec A\tan A}$

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

$=\frac{2\tan A}{2\sec A}$

$=\frac{\sin A}{\cos A}\times \cos A=\sin A$

Hence, proved.