Question

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

• A. $p$
• B. $\sqrt{p^2-q^2}$
• C. $\sqrt{p^2+q^2}\cos \theta$
• D. $noneofthese$

Answer 1

The correct option is A $p$

Given

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

$⟹\tan \theta =\frac{p}{\sqrt{q^2-p^2}}$

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

$⟹\sin \theta =\frac{p}{\sqrt{q^2}}$

$⟹q\sin \theta =p$