Question

lf $\tan \theta =\frac{p}{q}$ and $\theta =3\varphi (0<\theta <\frac{\pi }{2})$, then $\frac{p}{\sin \varphi }-\frac{q}{\cos \varphi }=$

• A. $2\sqrt{(p^2-q^2)}$
• B. $2\sqrt{(p^2+q^2)}$
• C. $\sqrt{(p^2+q^2)}$
• D. $3\sqrt{(p^2+q^2)}$

Answer 1

The correct option is B $2\sqrt{(p^2+q^2)}$

$\frac{p}{\sin \varphi }-\frac{q}{\cos \varphi }$
$=\frac{p\cos \varphi -q\sin \varphi }{\cos \varphi \sin \varphi }$
[$tan\theta =\frac{p}{q}$]
$\Rightarrow \cos \theta =\frac{q}{\sqrt{p^2+q^2}}$ and $\sin \theta =\frac{p}{\sqrt{p^2+q^2}}$ ]
$=\frac{p}{\sqrt{p^2+q^2}}\cos \varphi -\frac{q}{\sqrt{p^2+q^2}}\sin \varphi$
$=\sqrt{p^2+q^2}\frac{\sin \theta \cos \varphi -\cos \theta \sin \varphi }{\sin \varphi \cos \varphi }$
$=\frac{\sin (\theta -\varphi )}{\sin \varphi \cos \varphi }$ [$\theta =3\varphi$]
$=\sqrt{p^2+q^2}\frac{\sin (3\varphi -\varphi )}{\sin \varphi \cos \varphi }$
$=\sqrt{p^2+q^2}\frac{2\sin \varphi \cos \varphi }{\sin \varphi \cos \varphi }$
$=2\sqrt{p^2+q^2}$
Hence, option 'B' is correct.