Simplifying $\frac{6\tan x }{1-\tan^2 x}$ to $3 \tan 2x$

Question

How do I perform the following simplification? $$\frac{6\tan x }{1-\tan^2 x} \qquad\to\qquad 3 \tan 2x$$

I tried to take a common factor from denominator then use the laws of trigonometry, but I got nothing. Please explain the steps.

Answer 1

Simply note that the following identity holds

$$\tan 2x =\frac{2\tan x}{1-\tan^2 x}$$

which can be easily checked by the following

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