Using trigonometric identities to simply the following expression $\tan\frac{\pi}{5} + 2\tan\frac{2\pi}{5}+ 4\cot\frac{4\pi}{5}=\cot\frac{\pi}{5}$

Question

I know this is simple, but I just can't seem to get it... The question is to simplify:

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

To get: $\cot\frac{\pi}{5}$

How does one go about this?

Answer 1

\begin{align*} &\text{Using the identity}\\[6pt] &\qquad\cot(2u)=\frac{\cot^2(u)-1}{2\cot(u)}\\[10pt] &\tan(x)+2\tan(2x)+4\cot(4x)\\[4pt] =\;&\tan(x)+2\tan(2x)+4\frac{\cot^2(2x)-1}{2\cot(2x)}\\[4pt] =\;&\tan(x)+2\tan(2x)+2\frac{\cot^2(2x)-1}{\cot(2x)}\\[4pt] =\;&\tan(x)+\frac{2\tan(2x)\cot(2x)+2(\cot^2(2x)-1)}{\cot(2x)}\\[4pt] =\;&\tan(x)+\frac{2+2\cot^2(2x)-2)}{\cot(2x)}\\[4pt] =\;&\tan(x)+\frac{2\cot^2(2x))}{\cot(2x)}\\[4pt] =\;&\tan(x)+2\cot(2x)\\[4pt] =\;&\tan(x)+2\frac{\cot^2(x)-1}{2\cot(x)}\\[4pt] =\;&\tan(x)+\frac{\cot^2(x)-1}{\cot(x)}\\[4pt] =\;&\frac{\tan(x)\cot(x)+(\cot^2(x)-1)}{\cot(x)}\\[4pt] =\;&\frac{1+(\cot^2(x)-1)}{\cot(x)}\\[4pt] =\;&\frac{\cot^2(x)}{\cot(x)}\\[4pt] =\;&\cot(x)\\[4pt] \end{align*}

Answer 2

Add and subtract $\cot \frac {\pi}{5}$ ( insted of $pi/5$ u can keep $x$ )

$\implies (\cot \frac{\pi}{5} - \cot \frac{\pi}{5} + \tan \frac{\pi}{5} + 2 \tan \frac{2\pi}{5} + 4 cot\frac{\pi}5)$

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

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

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

$ =(\cot \frac{\pi}5 - 4\cot \frac{4\pi}5 + 4\cot\frac{4\pi}5)$

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

2nd question can also prove with this method :) ( answer is $Cot A$ )

Answer 3

Let $\theta = \frac{ \pi}{5}$ \begin{eqnarray*} \tan \theta + 2 \tan 2\theta + 4 \cot 4 \theta &=& \frac{\sin \theta }{\cos \theta} +\frac{2\sin 2\theta }{\cos 2\theta} +\frac{ 4\cos 4 \theta }{\sin 4 \theta} \\ &=& \frac{\sin \theta }{\cos \theta} +\frac{2\sin 2\theta }{\cos 2\theta} +\frac{4\cos 4 \theta }{2 \sin 2 \theta \cos 2 \theta } \\ &=& \frac{\sin \theta }{\cos \theta} +\frac{2\sin^2 2\theta +2\cos 4 \theta }{ \sin 2 \theta \cos 2 \theta } \\ &=& \frac{\sin \theta }{\cos \theta} +\frac{2\cos 2\theta }{ \sin 2 \theta } \\ &=& \frac{\sin \theta }{\cos \theta} +\frac{2\cos 2\theta }{2 \sin \theta \cos \theta } \\ &=& \frac{\sin^2 \theta +\cos 2\theta }{ \sin \theta \cos \theta } \\ &=& \frac{\cos \theta }{ \sin \theta }=\cot \theta \\ \end{eqnarray*}

Answer 4

Write $t:=\tan\frac{\pi}{5}$, so your expression is $$t+\frac{4t}{1-t^{2}}+\frac{1-6t^{2}+t^{4}}{t\left(1-t^{2}\right)}=\frac{t^{2}\left(1-t^{2}\right)+4t^{2}+1-6t^{2}+t^{4}}{t\left(1-t^{2}\right)}=\frac{1}{t}.$$