Question

Prove:

$\sec (x-y)=\frac{\sec x\sec y}{1+\tan x\tan y}$

Answer 1

Now,

$\sec (x-y)$

$=\frac{1}{\cos (x-y)}$

$=\frac{1}{(\cos x\cos y+\sin x.\sin y)}$

$=\frac{\sec x\sec y}{1+\tan x\tan y}$ [ Dividing the numerator and denominator by $\cos x.\cos y$]