Q:

$\sin 2x - \tan 2x = -\sin 2x\tan 2x$ trigonometric identity proof

I need to prove 

$$\sin 2x - \tan 2x = -\sin 2x\tan 2x$$

I tried simplifying 
$$
\sin 2x = 2\sin x\cos x;\quad  \tan 2x = \frac{2\tan x}{1-\tan^2x}.
$$
But it's so long and complicated that I feel I must've made a colossal mistake somewhere, or I am not using the correct equivalence. Or maybe I made a huge oversight from the beginning, and I don't need to simplify them in the first place.
Either way, i am stuck. Please help. Thanks.

A:

The identity is false at $x=\pi/6.$ Left side $-\sqrt{3}/2,$ right side $-3/2.$
Added: Suppose we want the left side to come out $-\tan 2x \cdot G,$ for some $G$ or other. After division of the sides by $\tan 2x$ this gives $G=1-\cos 2x.$ So as stated the identity would only hold provided $1-\cos 2x=\sin 2x.$ The latter is not an identity, one could solve it for the possible $x$ making it true. However it seems strikingly close (with major oversight) to the true identity $1-\cos^2 x=\sin^2 x,$ which leads me to guess that, IF someone were trying to cook up a simple identity exercise, and made this oversight, the "identity" of the post might be given as an exercise. Not to say there is any merit in this; I was just fooling around with the identity to see if maybe it was adjustable to a true identity somehow.
Another note: The identity could be "rescued" as
$$\sin(2x)-\tan(2x)=-2(\sin x)^2 \tan (2x).$$ With the common notation $\sin^2 x$ for the square of the sine, this means OP may have just been inexact on putting the exponent there, but it still needs that extra factor of $2.$

A:

What if $2x=n\pi+\dfrac\pi2$ where $n$ is any integer
Otherwise,
$$\sin2x+\sin2x\tan2x=\sin2x\cdot\dfrac{\cos2x+\sin2x}{\cos2x}=\tan2x(\cos2x+\sin2x)$$ whcih needs to be $$\tan2x$$ which will be true $\iff$
$$1=\cos2x+\sin2x=\dfrac{1-\tan^2x+2\tan x}{1+\tan^2x}$$
$$\iff\tan x(\tan x-1)=0$$

