Question

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

• A. $\cot \frac{\pi }{5}$
• B. $\cot \frac{2\pi }{5}$
• C. $\cot \frac{3\pi }{5}$
• D. $\cot \frac{4\pi }{5}$

Answer 1

Answer: (A)

Given,
$\theta =\frac{\pi }{5}=36^{\circ }$
$\tan \left(\frac{\pi }{5}\right)=\tan 36^{\circ }=\sqrt{5-2\sqrt{5}}$
$\tan 2\theta =\tan \left(\frac{2\pi }{5}\right)=\frac{2\tan 36^{\circ }}{1-{\tan }^{2}36^{\circ }}$
$=\frac{2\sqrt{5-2\sqrt{5}}}{1-(5-2\sqrt{5})}$
$=\frac{2\sqrt{5-2\sqrt{5}}}{2\sqrt{5}-4}=\frac{\sqrt{5-2\sqrt{5}}}{\sqrt{5}-2}$
Now, $\tan \left(\frac{4\pi }{5}\right)=\frac{2\tan \left(\frac{2\pi }{5}\right)}{1-{\tan }^2\frac{2\pi }{5}}=\frac{\frac{2\cdot \sqrt{5-2\sqrt{5}}}{\sqrt{5}-2}}{1-\frac{5-2\sqrt{5}}{(\sqrt{5}-2{)}^2}}$
$\therefore \cot \left(\frac{4\pi }{5}\right)=\frac{1-\frac{5-2\sqrt{5}}{(\sqrt{5}-2{)}^2}}{\frac{2\cdot \sqrt{5-2\sqrt{5}}}{\sqrt{5}-2}}$
$=\frac{(\sqrt{5}-2{)}^2-5+2\sqrt{5}}{2(\sqrt{5}-2)(\sqrt{5-2\sqrt{5}})}=\frac{5+4-4\sqrt{5}-5+2\sqrt{5}}{2(\sqrt{5}-2)(\sqrt{5-2\sqrt{5}})}$
$=\frac{4-2\sqrt{5}}{2(\sqrt{5}-2)(\sqrt{5-2\sqrt{5}})}=\frac{(2-\sqrt{5})}{(\sqrt{5}-2)(\sqrt{5-2\sqrt{5}})}$
$=\frac{-1}{\sqrt{5-2\sqrt{5}}}$
Now, $\tan \frac{\pi }{5}+2\tan \frac{2\pi }{5}+4\cot \frac{4\pi }{5}$
$=\sqrt{5-2\sqrt{5}}+2\cdot \frac{\sqrt{5-2\sqrt{5}}}{\sqrt{5}-2}-\frac{4}{\sqrt{5-2\sqrt{5}}}$
$=\sqrt{5-2\sqrt{5}}+2\frac{\sqrt{5-2\sqrt{5}}}{\sqrt{5}-2}\times \frac{\sqrt{5}+2}{\sqrt{5}+2}-\frac{4}{\sqrt{5-2\sqrt{5}}}$
$=\sqrt{5-2\sqrt{5}}+2(\sqrt{5}+2\sqrt{5-2\sqrt{5}}-\frac{4}{\sqrt{5-2\sqrt{5}}}$
$=\sqrt{5-2\sqrt{5}}+2(\sqrt{5}+2)\sqrt{5-2\sqrt{5}}-\frac{4}{\sqrt{5-2\sqrt{5}}}$
$=\frac{5-2\sqrt{5}+2(\sqrt{5}+2)(5-2\sqrt{5})-4}{\sqrt{5-2\sqrt{5}}}$
$=5-2\sqrt{5}+(2\sqrt{5}+4)(5-2\sqrt{5})-4$
$=\frac{5-2\sqrt{5}+10\sqrt{5}-20+20-8\sqrt{5}-4}{\sqrt{5-2\sqrt{5}}}$
$=\frac{1}{\sqrt{5-2\sqrt{5}}}=1\cdot 37=\cot \frac{\pi }{5}=\cot 36^{\circ }$