Question

If $\frac{1-10i\cos \theta }{1-10\sqrt{3}i\sin \theta }$ is purely real, then one of the values of $\theta$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

$\frac{1-10i\cos \theta }{1-10\sqrt{3}i\sin \theta }\times \frac{1+10\sqrt{3}i\sin \theta }{1+10\sqrt{3}i\sin \theta }$
$=\frac{1+10\sqrt{3}i\sin \theta -10i\cos \theta +100\sqrt{3}\sin \theta \cos \theta }{1+300{\sin }^2\theta }$
$=\frac{1+100\sqrt{3}\sin \theta \cos \theta }{1+300{\sin }^2\theta }+i\frac{10\sqrt{3}\sin \theta -10\cos \theta }{1+300{\sin }^2\theta }$
purely real so imaginary part $=0$
$\Rightarrow \frac{10\sqrt{3}\sin \theta -10\cos \theta }{1+300{\sin }^2\theta }=0$
$\Rightarrow 1+300{\sin }^2\theta \ne 0$
So,$10\sqrt{3}\sin \theta -10\cos \theta =0$
$\Rightarrow \tan \theta =\frac{10}{10\sqrt{3}}$
$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}=\tan \left(\frac{\pi }{6}\right)$
$\Rightarrow \theta =\frac{\pi }{6}$