I need to use the fact that $\tan 2x=\sin2x \ /\cos2x$ to prove that: $$\tan 2x=\frac{2\tan x}{1-\tan^2x}$$ I don't know where to start. Please help or hint. Thanks in advance.


Use the identities: $$ \cos{(2x)}=\cos^2{(x)}-\sin^2{(x)}$$ and $$ \sin{(2x)}=2\sin(x)\cos(x) $$

since you'll get: $$ \frac{2\sin(x)\cos(x)}{\cos^2{(x)}-\sin^2{(x)}} $$ can you take it from there?

  • $\begingroup$ Yes thank you very much for your quick and helpful reply $\endgroup$ – igeer12 May 9 '14 at 0:44
  • $\begingroup$ @igeer12 No worries, glad to help! $\endgroup$ – Jay May 9 '14 at 0:46

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