Question

Given: $\sin \Theta =\frac{p}{q}$ , find $\cos \Theta +\sin \Theta$ in terms of $p$ and $q$,

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

Answer 1

Answer: (C)

$\frac{p+\sqrt{q^2-p^2}}{q}$

Given, $\sin \Theta =\frac{p}{q}$
$\sin \Theta =\frac{P}{H}=\frac{p}{q}$
Now, by Pythagoras Theorem,
$H^2=P^2+B^2$
$q^2=p^2+B^2$
$B=\sqrt{q^2-p^2}$
$\sin \Theta +\cos \Theta =\frac{P}{H}+\frac{B}{H}$
= $\frac{p}{q}+\frac{\sqrt{q^2-p^2}}{q}$
= $\frac{p+\sqrt{q^2-p^2}}{q}$