Prove $\tan\frac\pi{16}+2\tan\frac\pi8+4=\cot\frac{\pi}{16}$

Question

Prove that $\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\cot\dfrac{\pi}{16}$

My Attempt \begin{align} &\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\dfrac{1}{\cot\dfrac{\pi}{16}}+\dfrac{2}{\cot\dfrac{\pi}{8}}+4\\ &=\dfrac{1}{\cot\dfrac{\pi}{16}}+2\dfrac{2\cot\dfrac{\pi}{16}}{\cot^2\dfrac{\pi}{16}-1}+4=\dfrac{\cot^2\dfrac{\pi}{16}-1+4\cot^2\dfrac{\pi}{16}+4\cot^3\dfrac{\pi}{16}-\cot\dfrac{\pi}{16}}{\cot\dfrac{\pi}{16}(\cot^2\dfrac{\pi}{16}-1)}\\ &= \end{align}

I don't think its going anywhere with my attempt, Is there an easy way to prove this ?

I have checked a similar post Reducing $\tan\frac{\pi}{16} + 2\tan\frac{\pi}{8}+4$ to $\cot\frac{\pi}{16}$, but as it was a multiple choice question hope there \d be any direct way to solve this.

Answer 1

Given $$\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\cot\dfrac{\pi}{16}$$

Now $$\cot \theta-\tan\theta=\dfrac{\cos\theta}{\sin\theta}-\dfrac{\sin\theta}{\cos\theta}=\dfrac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}=\dfrac{\cos2\theta}{\dfrac12\sin2\theta}=2\cot2\theta$$

Since $\cot\theta-\tan\theta=2\cot2\theta\ \ $ we get

$$\cot\theta-2\cot2\theta=\tan\theta$$

Now

$\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\cot\dfrac{\pi}{16}-2\cot\dfrac{\pi}{8}+2\left(\cot\dfrac{\pi}{8}-2\cot\dfrac{\pi}{4}\right)+4$

$$=\cot\dfrac{\pi}{16}-4\cot\dfrac{\pi}{4}+4$$$$=\cot\dfrac{\pi}{16}-4+4$$ $$=\cot\dfrac{\pi}{16}$$

Therefore,

$$\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\cot\dfrac{\pi}{16}$$