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KEAM 2017 Math Paper

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Question : 86 of 120

Marks: +1, -0

If

a=e^{\theta},

\frac{1+a}{1-a}

is equal to

$\cot \frac{\theta}{2}$
tan θ
$i \cot \frac{\theta}{2}$
$i \tan \frac{\theta}{2}$
2 tan θ

Solution:

We have

a=e^{i \theta}

=\cos \theta + i \sin \theta

Now,

\; \frac{1+a}{1-a} = \frac{1+(\cos \theta + i \sin \theta)}{1-(\cos \theta + i \sin \theta)}

= \frac{(1+\cos \theta) + i \sin \theta}{(1-\cos \theta) - i \sin \theta}

= \frac{2 \cos^{2} \frac{\theta}{2} + i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^{2} \frac{\theta}{2} - i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}

= \frac{2 \cos \frac{\theta}{2} \left[\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}\right]}{2 \sin \frac{\theta}{2} \left[\sin \frac{\theta}{2} - i \cos \frac{\theta}{2}\right]}

= \frac{\cot \frac{\theta}{2} \left[\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}\right]}{-i \left[\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}\right]}

= \frac{\cot \frac{\theta}{2}}{-i}

= i \cot \frac{\theta}{2}

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