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