Question

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

Answer 1

$$\begin{gathered} \mathrm{LHS}: \\ \frac{a\sin x-b\cos x}{a\sin x+b\cos x} \\ \mathrm{Dividing}\mathrm{by}\mathit{}b\cos x: \\ =\frac{{\displaystyle \frac{a\tan x}{b}}-1}{{\displaystyle \frac{a\tan x}{b}}+1} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\tan x \\ =\frac{a^2-b^2}{a^2+b^2} \\ =\mathrm{RHS} \end{gathered}$$
Hence proved.