Question:
If $\tan x=\frac{a}{b}$, show that $\frac{a \sin x-b \cos \mathrm{x}}{a \sin x+b \cos x}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$.
Solution:
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.
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