Question

If \[\tan x = \frac{a}{b},\] show that

\[\frac{a \sin x - b \cos x}{a \sin x + b \cos x} = \frac{a^2 - b^2}{a^2 + b^2}\]

Answer 1

LHS: 
\[\frac{a\sin x - b\cos x}{a\sin x + b\cos x}\]
Dividing by \[b\cos x: \]
\[ = \frac{\frac{a\tan x}{b} - 1}{\frac{a\tan x}{b} + 1}\]
Substituting the value of \[\tan x\]
\[ = \frac{a^2 - b^2}{a^2 + b^2}\]
 = RHS
Hence proved.