Question

The value of $\tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)$ is

• A. $\cot \left(\frac{\mathrm{\pi }}{5}\right)$
• B. $\cot \left(\frac{2\mathrm{\pi }}{5}\right)$
• C. $\cot \left(\frac{4\mathrm{\pi }}{5}\right)$
• D. $\cot \left(\frac{3\mathrm{\pi }}{5}\right)$

Answer 1

The correct option is A

$\cot \left(\frac{\mathrm{\pi }}{5}\right)$

Explanation for the correct option

The given trigonometric expression: $\tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)$.

$$\begin{gathered} \tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)=\tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(2\times \frac{2\mathrm{\pi }}{5}\right) \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)=\tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+\frac{4}{\tan \left(2\times \frac{2\mathrm{\pi }}{5}\right)}\left[\because \cot \left(\theta \right)=\frac{1}{\tan \left(\theta \right)}\right] \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)=\tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+\frac{4}{{\displaystyle \frac{2\tan \left(\frac{2\mathrm{\pi }}{5}\right)}{1-{\tan }^2\left({\displaystyle \frac{2\mathrm{\pi }}{5}}\right)}}}\left[\because \tan \left(2\theta \right)=\frac{2\tan \left(\theta \right)}{1-{\tan }^2\left(\theta \right)}\right] \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)=\tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+\frac{2-2{\tan }^2\left(\frac{2\mathrm{\pi }}{5}\right)}{{\displaystyle \tan \left(\frac{{\displaystyle 2\mathrm{\pi }}}{{\displaystyle 5}}\right)}} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)=\tan \left(\frac{\mathrm{\pi }}{5}\right)+\frac{2{\tan }^2\left(\frac{2\mathrm{\pi }}{5}\right)+2-2{\tan }^2\left(\frac{2\mathrm{\pi }}{5}\right)}{{\displaystyle \tan \left(2\times \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 5}}\right)}} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)=\tan \left(\frac{\mathrm{\pi }}{5}\right)+\frac{2}{{\displaystyle \tan \left(2\times \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 5}}\right)}} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)=\tan \left(\frac{\mathrm{\pi }}{5}\right)+\frac{2}{{\displaystyle \left(\frac{{\displaystyle 2\tan \left(\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 5}}\right)}}{{\displaystyle 1-{\tan }^2\left(\frac{\mathrm{\pi }}{5}\right)}}\right)}} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)=\tan \left(\frac{\mathrm{\pi }}{5}\right)+\frac{1-{\tan }^2\left(\frac{\mathrm{\pi }}{5}\right)}{{\displaystyle \tan \left(\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 5}}\right)}} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)=\frac{{\tan }^2\left(\frac{\mathrm{\pi }}{5}\right)+1-{\tan }^2\left(\frac{\mathrm{\pi }}{5}\right)}{{\displaystyle \tan \left(\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 5}}\right)}} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)=\frac{1}{{\displaystyle \tan \left(\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 5}}\right)}} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{5}\right)+2\tan \left(\frac{2\mathrm{\pi }}{5}\right)+4\cot \left(\frac{4\mathrm{\pi }}{5}\right)=\cot \left(\frac{\mathrm{\pi }}{5}\right) \end{gathered}$$

Hence, option A is correct .