Question

If $pcot\theta =\sqrt{q^2-p^2}$, then the value of $sin\theta$ is _________.

• A. $\frac{p}{q}$
• B. $\frac{q}{2p}$
• C. $\frac{q}{p}$
• D. 0

Answer 1

The correct option is A $\frac{p}{q}$

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

$\therefore cot\theta =\frac{\sqrt{q^2-p^2}}{p}$

We know that, $cose{c}^2\theta =1+co{t}^2\theta$.
By substituting the value of $cot\theta =\frac{\sqrt{q^2-p^2}}{p}$ in the trigonometric identity we get,
$cose{c}^2\theta =1+\frac{q^2-p^2}{p^2}=\frac{q^2}{p^2}$

So, $cose{c}^2\theta =\frac{q^2}{p^2}$

By taking the square root on both the sides,
$cosec\theta =\frac{q}{p}$
And, we know that, $cosec\theta =\frac{1}{sin\theta }$.
$\therefore sin\theta =\frac{p}{q}$