Question

Show that $\tan (\frac{\pi }{3}+x)$ $\tan (\frac{\pi }{3}-x)=\frac{2\cos 2x+1}{2\cos 2x-1}$.

Answer 1

Simplifying the LHS of $\tan \left(\frac{\pi }{3}+x\right)\tan \left(\frac{\pi }{3}-x\right)=\frac{2\cos 2x+1}{2\cos 2x-1}$

$\tan \left(\frac{\pi }{3}+x\right)\tan \left(\frac{\pi }{3}-x\right)=\left(\frac{\tan \frac{\pi }{3}+\tan x}{1-\tan \frac{\pi }{3}\tan x}\right)\left(\frac{\tan \frac{\pi }{3}-\tan x}{1+\tan \frac{\pi }{3}\tan x}\right)$

$=\frac{{\left(\tan \frac{\pi }{3}\right)}^2-{\left(\tan x\right)}^2}{{\left(1\right)}^2-{\left(\tan \frac{\pi }{3}\tan x\right)}^2}$

$=\frac{{\left(\sqrt{3}\right)}^2-{\tan }^{2}x}{1-{\left(\sqrt{3}\tan x\right)}^2}$

$=\frac{3-\frac{{\sin }^{2}x}{{\cos }^{2}x}}{1-3\frac{{\sin }^{2}x}{{\cos }^{2}x}}$

$=\frac{3{\cos }^{2}x-{\sin }^{2}x}{{\cos }^{2}x-3{\sin }^{2}x}$

$=\frac{3{\cos }^{2}x-\left(1-{\cos }^{2}x\right)}{{\cos }^{2}x-3\left(1-{\cos }^{2}x\right)}$

$=\frac{3{\cos }^{2}x-1+{\cos }^{2}x}{{\cos }^{2}x-3+3{\cos }^{2}x}$

$=\frac{4{\cos }^{2}x-1}{4{\cos }^{2}x-3}$

Simplifying the RHS of $\tan \left(\frac{\pi }{3}+x\right)\tan \left(\frac{\pi }{3}-x\right)=\frac{2\cos 2x+1}{2\cos 2x-1}$

$\frac{2\cos 2x+1}{2\cos 2x-1}=\frac{2\left(2{\cos }^{2}x-1\right)+1}{2\left(2{\cos }^{2}x-1\right)-1}$

$=\frac{4{\cos }^{2}x-2+1}{4{\cos }^{2}x-2-1}$

$=\frac{4{\cos }^{2}x-1}{4{\cos }^{2}x-3}$

Therefore, $LHS=RHS$.

Hence proved.