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

Question

Aicharya · Accepted Answer

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

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

Solution