Question

If $\tan \theta =p/q$, show that
$\frac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }=\frac{p^2-q^2}{p^2+q^2}$.

Answer 1

$\frac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }=\frac{p\tan \theta -q}{p\tan \theta +q}$, on dividing by $\cos \theta$
$\frac{p\left(p/q\right)-q}{p\left(p/q\right)+q}=\frac{p^2-q^2}{p^2+q^2}$ $(\because \tan \theta =\frac{p}{q})$