Question

If $qcos\Theta =\sqrt{q^2-p^2},$ prove that q sin $\Theta$=p.

Answer 1

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

To prove:- $q\sin \theta =p$

Proof:-

$q\cos \theta =\sqrt{q^2-p^2}\left(\text{Given}\right)$

$\Rightarrow \cos \theta =\frac{\sqrt{q^2-p^2}}{q}.....\left(1\right)$

As we know that,

$\cos \theta =\frac{\text{Base}}{\text{Hypotenuse}}.....\left(2\right)$

On comparing ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\text{Base}=\sqrt{q^2-p^2}$

$\text{Hypotenuse}=q$

Now applying pythagoras theorem,

${\text{Hypotenuse}}^2={\text{Base}}^2+{\text{Perpendicular}}^2$

$\Rightarrow {\text{Perpendicular}}^2=q^2-{\left(\sqrt{q^2-p^2}\right)}^2$

$\Rightarrow {\text{Perpendicular}}^2=q^2-(q^2-p^2)$

$\Rightarrow {\text{Perpendicular}}^2=q^2-q^2+p^2$

$\Rightarrow \text{Perpendicular}=\sqrt{p^2}=p$

Now again as we know that,

$\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

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

$\Rightarrow q\sin \theta =p$

Hence proved.