Question

(sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x

Answer 1

I'm assuming you need to show that the equation is true.
First off, use what you know (or are learning) about trigonometric identities. We see that the simplified side has only terms of cos, so it will be beneficial to re-write the left side in terms of just sin(x) and cos(x). So..
\[\sin x \left( \frac{sin x}{cos x} * \cos x - \frac{cos x}{sin x} * \cos x \right)\]
Next, you'll want to simplify terms as best you can. Start by combining and canceling like terms inside the parenthesis:
\[\sin x \left( \sin x - \frac{cos^2x}{sinx} \right)\]
Next distribute the sin(x) into the parenthesis:
\[\left( sin^2x - cos^2x\right)\]
Next step isn't too obvious unless you know your double angle formulas pretty well. If you look them up, you'll find that \[\cos^2x - \sin^2x = 2\cos^2x -1\]
Well, on the left side we have:
\[\sin^2x - \cos^2x\] instead of \[\cos^2x - \sin^2x\]
so we can just reorder the usual double angle identity also (just invert it) to get
\[\sin^2x - cos^2x = 1 - 2cos(2x)\]

Hope this helps (and hopefully I didn't omit anything or make a mistake lol)!