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.

  • $\begingroup$ math.stackexchange.com/questions/2305793/… $\endgroup$ – lab bhattacharjee Nov 12 '18 at 6:12
  • $\begingroup$ @labbhattacharjee thanx. but actually i don't want to use that formula $\endgroup$ – ss1729 Nov 12 '18 at 6:20
  • 1
    $\begingroup$ @ss1729 why not? $2\cot 2x=\cot x-\tan x$ is what you get by taking reciprocal of the double formula for tan. $\endgroup$ – user10354138 Nov 12 '18 at 6:36
  • $\begingroup$ @ss1729, The accepted answer has used that formula only, right? $\endgroup$ – lab bhattacharjee Nov 12 '18 at 10:43
  • $\begingroup$ @labbhattacharjee thats right, and thanx. u are right i think i realise, using that formula is the best way to solve this problem. $\endgroup$ – ss1729 Nov 12 '18 at 13:50

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




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




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