Respuesta :
[tex]\bf sin(x)~~[ tan(x)cos(x)-cot(x)cos(x) ]~~=~~1-2cos^2(x) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(x)~~\left[\cfrac{sin(x)}{\underline{cos(x)}}\cdot \underline{cos(x)}~~-~~\cfrac{cos(x)}{sin(x)}\cdot cos(x) \right] \\\\\\ sin(x)~~\left[\cfrac{sin(x)}{1}~~-~~\cfrac{cos^2(x)}{sin(x)} \right]\implies \underline{sin(x)~}~\left[\cfrac{sin^2(x)-cos^2(x)}{\underline{sin(x)}} \right] \\\\\\ sin^2(x)-cos^2(x)\implies -[cos^2(x)-sin^2(x)] \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \textit{Pythagorean Identities} \\\\ sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta) \\\\ 1+cot^2(\theta)=csc^2(\theta) \\\\ 1+tan^2(\theta)=sec^2(\theta) \\\\[-0.35em] ~\dotfill\\\\ -[cos^2(x)-sin^2(x)]\implies -[cos^2(x)-[1-cos^2(x)]] \\\\\\ -[cos^2(x)-1+cos^2(x)]\implies -[2cos^2(x)-1]\implies 1-2cos^2(x)[/tex]
It is proved that the left side equals the right side: (sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x and this will be proved by using the trigonometry properties.
Given :
(sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x
To prove that the left-hand side equals the right-hand side, take the left-hand part.
[tex]\rm LHS = sinx (tanx\;cosx-cotx\;cosx)[/tex]
Now, write [tex]\rm tanx = \dfrac{sinx}{cosx}[/tex] and [tex]\rm cotx = \dfrac{cosx}{sinx}[/tex] in the above equation.
[tex]\rm LHS= sinx (\dfrac{sinx}{cosx}\;cosx-\dfrac{cosx}{sinx}\;cosx)[/tex]
Now, further simplify the above eqaution.
[tex]\rm LHS= (sinx)^2-(cosx)^2[/tex]
Now, apply the property [tex]\rm sin^2x-cos^2x=1-2cos2x[/tex].
[tex]\rm LHS = 1-2cos2x = RHS[/tex]
For more information, refer to the link given below:
https://brainly.com/question/10644258
