Question

If $\frac{1-10 i \cos \theta}{1-10 \sqrt{3} i \sin \theta}$ 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)

$\frac{1-10 i \cos \theta}{1-10 \sqrt{3} i \sin \theta} \times \frac{1+10 \sqrt{3} i \sin \theta}{1+10 \sqrt{3} i \sin \theta}$
$=\frac{1+10 \sqrt{3} i \sin \theta-10 i \cos \theta+100 \sqrt{3} \sin \theta \cos \theta}{1+300 \sin ^{2} \theta}$
$=\frac{1+100 \sqrt{3} \sin \theta \cos \theta}{1+300 \sin ^{2} \theta}+i \frac{10 \sqrt{3} \sin \theta-10 \cos \theta}{1+300 \sin ^{2} \theta}$
purely real so imaginary part $=0$
$\Rightarrow \, \frac{10 \sqrt{3} \sin \theta-10 \cos \theta}{1+300 \sin ^{2} \theta}=0$
$\Rightarrow \, 1+300 \sin ^{2} \theta \neq 0$
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}$