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AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

Section: Mathematics

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Question : 21 of 160

Marks: +1, -0

tan

\frac{\pi}{5}

\frac{2\pi}{5}

\frac{4\pi}{5}

=

cot $\frac{\pi}{5}$
cot $\frac{2\pi}{5}$
cot $\frac{3\pi}{5}$
cot $\frac{4\pi}{5}$

Solution:

It is given that,

\theta = \frac{\pi}{5}

=36^{\circ}

This implies,

\tan \left( \frac{\pi}{5} \right) = \tan 36^{\circ}

= \sqrt{5-2\sqrt{5}}

Now,

\tan 2\theta = \tan \left( \frac{2\pi}{5} \right)

= \frac{2 \tan 36^{\circ}}{1 - \tan^{2} 36^{\circ}}

= \frac{2 \sqrt{5-2\sqrt{5}}}{1-5-2\sqrt{5}}

= \frac{\sqrt{5} - 2\sqrt{5}}{\sqrt{5} - 2}

Then,

\tan \left( \frac{4\pi}{5} \right) = \frac{ \frac{2 \sqrt{5-2\sqrt{5}}}{\sqrt{5}-2} }{ 1 - \frac{5-2\sqrt{5}}{\sqrt{5}-2} }

\cot \left( \frac{4\pi}{5} \right) = \frac{ 1 - \frac{5-2\sqrt{5}}{\sqrt{5}-2} }{ \frac{2\sqrt{5-2\sqrt{5}}}{\sqrt{5}-2} }

= \frac{(2-\sqrt{5})}{(\sqrt{5}-2)(\sqrt{5-2\sqrt{5}})}

= \frac{-1}{\sqrt{5-2\sqrt{5}}}

Now,

\tan \frac{\pi}{5} + 2 \tan \frac{2\pi}{5} + 4 \cot \frac{4\pi}{5}

= \sqrt{5-2\sqrt{5}} + 2 \left( \frac{\sqrt{5-2\sqrt{5}}}{\sqrt{5}-2} \right)

\left( -\frac{4}{\sqrt{5-2\sqrt{5}}} \right)

= -1.37

= \cot \frac{\pi}{5}

= \cot 36^{\circ}

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