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$$\text{The value of}\quad\tan\frac{\pi}{16} + 2\tan\frac{\pi}{8} +4 \quad\text{is equal to _______.}$$ (Answer: $\cot\frac{\pi}{16}$)

I solved the question by the identity $$\tan \phi = \cot\phi-2\cot 2\phi$$ and got the right answer. However, I want to get some other way so that I can solve the question using basic expansions of tangent, rather than using such an uncommon identity (which I had to look up in the book).

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    $\begingroup$ @Blue Thanks dude. $\endgroup$ – Harsh Sharma Jun 5 '16 at 4:51
  • $\begingroup$ Probably there is no simple formula $\endgroup$ – Archis Welankar Jun 5 '16 at 5:40
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Too long for a comment so writing it here . Write $4=4.\tan(\frac{\pi}{4})$ so we get an interesting GP in angles and also in the outer numbers ie the sum becomes $\tan(\frac{\pi}{16})+2\tan(2\frac{\pi}{16})+2^2\tan(2^2\frac{\pi}{16})$. Now as there is a very unnatural formula for it which goes like $\tan(x)+2\tan(2x)+4\tan(4x)=\cot(x)-8\cot(8x)$ As you wanted to use basic identities a good proof is given here .https://googleweblight.com/?lite_url=https://in.answers.yahoo.com/question/index?qid%3D20151113194901AAbOr9f&ei=uDY1i-ra&lc=en-IN&s=1&m=746&host=www.google.co.in&ts=1465105675&sig=APY536xf2Z18ttOg5VBlpB0ZtaXJyvsv8A

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  • $\begingroup$ Very nice and cool too. Never thought anything like this could have been done. $\endgroup$ – Harsh Sharma Jun 5 '16 at 6:44
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    $\begingroup$ You are welcome ;) $\endgroup$ – Archis Welankar Jun 5 '16 at 7:07

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