# Reducing $\tan\frac{\pi}{16} + 2\tan\frac{\pi}{8} +4$ to $\cot\frac{\pi}{16}$

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

• @Blue Thanks dude. – Harsh Sharma Jun 5 '16 at 4:51
• Probably there is no simple formula – Archis Welankar Jun 5 '16 at 5:40

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