Trig identity is this impossiable

Sin2x - tan2x = -sin2x tan2x

They are not exponents 2 they are numbers infront of x

Note that you can re-write tan(2x) as sin(2x)/cos(2x). Then:
LHS = sin(2x) - tan(2x)
= sin(2x) - sin(2x)/cos(2x)
= sin(2x)[1 - 1/cos(2x)]
= sin(2x)[cos(2x) - 1]/cos(2x), by getting a common denominator within brackets
= [sin(2x)/cos(2x)][cos(2x) - 1], by re-arranging
= tan(2x)[cos(2x) - 1].

At this point, the only way to proceed further is to find a way to show that:
cos(2x) - 1 = -sin(2x),

which is obviously false. Therefore, this is not an identity.

I hope this helps!

Sin 2x = 2 sin x cos x, tan 2x = (2 tan x)/(1 - tan²x)

Also tan x = sin x/cos x, and tan²x = sin²x/cos²x

sin 2x - tan 2x = -sin 2x tan 2x
2 sin x cos x - (2 tan x)/(1-tan²x) = (-4 sin x cos x tan x)/(1 - tan²x)
2 sin x cos x = (2 tan x - 4 sin x cos x tan x)/(1 - tan²x)

Multiply by 1 - tan²x to eliminate the fraction.

2 sin x cos x - 2 sin x cos x tan²x = 2 tan x - 4 sin x cos x tan x

Divide across by 2 and substitute for tan x and tan²x (doesn't look elegant, but it works)

sin x cos x - (sin x cos x sin²x)/cos²x = sin x/cos x - 2 sin x cos x sin x/cos x

Simplify.

sin x cos x - sin³x/cos x = sin x/cos x - 2 sin²x

Multiply by cos x

sin x cos²x - sin³x = sin x - 2 sin²x cos x

Substitute 1-sin²x for cos²x

sin x(1 - sin²x) - sin³x = sin x - 2 sin²x cos x
sin x - sin³x - sin³x = sin x - 2 sin²x cos x

Simplify again.

-2 sin³x = -2 sin²x cos x
sin³x = sin²x cos x

Divide by sin²x

sin x = cos x

This solves at x = 45° and x = 235°

I have tried to prove it and found it impossible. If we substitute different values of X LHS and RHS are not equal.

it gets easier