Question

If $\tan \theta =\frac{p}{q}$ and $\theta =3\varphi (0<\theta <\frac{\pi }{2})$, prove
that $\frac{p}{\sin \varphi }-\frac{q}{\cos \varphi }=2\sqrt{(p^2+q^2)}$

Answer 1

From given relation
$\frac{\sin \theta }{p}=\frac{\cos \theta }{q}=\sqrt{\frac{{\sin }^2\theta +{\cos }^2\theta }{p^2+q^2}}=\frac{1}{\sqrt{p^2+q^2}}$
Now putting for $p$ in the $q$ in the $L.H.S$., we have
$L.H.S=\sqrt{(p^2+q^2)}[\frac{\sin \theta }{\sin \varphi }-\frac{\cos \theta }{\cos \varphi }]=\sqrt{(p^2+q^2)}\frac{\sin (\theta -\varphi )}{\sin \varphi \cos \varphi }$
$=\sqrt{(p^2+q^2)}.2\frac{\sin (3\varphi -\varphi )}{\sin 2\varphi }=\sqrt{p^2+q^2}\because \theta =3\varphi$