Question

Prove that (tanx-1)(sin 2x-2 cos²x) = 2(1-2 sinxcos.x)​

Answer 1

`( ~~ \tan(x)-1 ~~ )( ~~ \sin(2x)-2\cos^2(x) ~~ )~~ = ~~2( ~~ 1-2\sin(x)\cos(x) ~~ ) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \tan(x)-1\implies \cfrac{\sin(x)}{\cos(x)}-1\implies \boxed{\cfrac{\sin(x)-\cos(x)}{\cos(x)}} \\\\[-0.35em] ~\dotfill\\\\ \sin(2x)-2\cos^2(x)\implies \boxed{2\sin(x)\cos(x)-2\cos^2(x)} \\\\[-0.35em] ~\dotfill\\\\ \left[ \cfrac{\sin(x)-\cos(x)}{\cos(x)} \right] ~~ [2\sin(x)\cos(x)-2\cos^2(x)]`

`\left[ \cfrac{\sin(x)-\cos(x)}{\cos(x)} \right]2\sin(x)\cos(x)~~ + ~~\left[ \cfrac{\sin(x)-\cos(x)}{\cos(x)} \right][-2\cos^2(x)] \\\\\\ \left[ \sin(x)-\cos(x) \right]\cdot 2\sin(x)~~ + ~~\left[ \sin(x)-\cos(x) \right][-2\cos(x)] \\\\\\ 2\sin^2(x)-2\sin(x)\cos(x)~~ + ~~[-2\sin(x)\cos(x)+2\cos^2(x)]`

`2\sin^2(x)-4\sin(x)\cos(x)+2\cos^2(x)\implies 2[\sin^2(x)-2\sin(x)\cos(x)+\cos^2(x)] \\\\\\ 2[\sin^2(x)+\cos^2(x)-2\sin(x)\cos(x)]\implies {\Large \begin{array}{llll} 2( ~~ 1-2\sin(x)\cos(x) ~~ ) \end{array}`