Question

Prove esta identidade:

$sec x - tg x = \frac{1}{secx + tgx}$

Answer 1

$\sec x-\tan x=\dfrac{1}{\sec x+\tan x}\\\\ \dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x}=\dfrac{1}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}}\\\\ \dfrac{1-\sin x}{\cos x}=\dfrac{1}{~~\frac{1+\sin x}{\cos x}~~}\\\\\\ (1-\sin x)(\dfrac{1+\sin x}{\cos x})=\cos x\\\\ \dfrac{1-\sin^2x}{\cos x}=\cos x\\\\ \dfrac{\cos^2x}{\cos x}=\cos x\\\\ \cos x=\cos x\Longrightarrow OK!~~\blacksquare$