Question

If $q\cos \theta =\sqrt{q^2-p^2}$ then prove $q\sin \theta =p$

Answer 1

$q\cos \theta =\sqrt{q^2-p^2}$

Squaring on both sides

$q^2{\cos }^2\theta =q^2-p^2\Rightarrow {\cos }^2\theta =\frac{q^2-p^2}{q^2}$

we known that

${\cos }^2\theta +{\sin }^2\theta =1$

substituting value pf ${\cos }^2\theta$

$\frac{q^2-p^2}{q^2}+{\sin }^2\theta =1$

${\sin }^2\theta =1-\left(\frac{q^2-p^2}{q^2}\right)$

${\sin }^2\theta =\frac{p^2}{q^2}$

$\sin \theta =\frac{p}{q}$

$q\sin \theta =p$

Hence, proved