Simplifying $\frac{6\tan x }{1-\tan^2 x}$ to $3 \tan 2x$
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.
$\endgroup$1 Answer
$\begingroup$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
- $\tan{2x}=\frac{\sin {2x}}{\cos{2x}}$
- $\sin{2x}=2\sin x\cos x$
- $\cos{2x}=\cos^2{x}-\sin^2{x}$
