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TS EAMCET 4 May 2018 Shift 2 Solved Paper

Section: Mathematics

Question : 7 of 160

Marks: +1, -0

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

\theta

Solution:

Consider the expression,

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

Rationalize the above,

\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}

The imaginary part will be zero for pure real value so,

\frac{10 \sqrt{3} \sin \theta - 10 \cos \theta}{1+300 \sin^2 \theta}=0

10 \sqrt{3} \sin \theta - 10 \cos \theta=0

\tan \theta = \frac{1}{\sqrt{3}}

\theta = \frac{\pi}{6}

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