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