Question

Prove that :
$\frac{\tan 3x}{\tan x}=\frac{2\cos 2x+1}{2\cos 2x-1}$

Answer 1

$\frac{tan3x}{tanx}=\frac{sin3x/cos3x}{sinx/cosx}$

$=\frac{(3sinx-4si{n}^{3}x)/(4co{s}^{3}x-3cosx)}{sinx/cosx}$

$=\frac{3-4si{n}^{2}x}{4co{s}^{2}x-3}=\frac{3-4(1-co{s}^{2}x)}{4co{s}^{2}x-3}$

$=\frac{3-4+4co{s}^{2}x}{4co{s}^{2}x-3}=\frac{4co{s}^{2}x-1}{4co{s}^{2}x-3}$

$=\frac{2\left(2co{s}^{2}x\right)-1}{2cos2x-1}=\frac{2(1+cos2x)-1}{2cos2x-1}$

$=\frac{2+2co{s}^{2}x-1}{2co{s}^{2}x-1}=\frac{2co{s}^{2}x+1}{2co{s}^{2}x-1}$

Hence, proved.