Question

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

Answer 1

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