Question

sin(x)/1-cos(x) - 1-cos(x)/sin(x) = 2cot(x)

Answer 1

let's recall the pythagorean identity that sin²(θ) + cos²(θ) = 1.

$\bf \cfrac{sin(x)}{1-cos(x)}-\cfrac{1-cos(x)}{sin(x)}=2cot(x) \\\\[-0.35em] ~\dotfill\\\\ \cfrac{sin(x)}{1-cos(x)}-\cfrac{1-cos(x)}{sin(x)}\implies \cfrac{[sin(x)]^2-[1-cos(x)]^2}{sin[1-cos(x)]} \\\\\\ \cfrac{[sin^2(x)]-[1^2-2cos(x)+cos^2(x)]}{sin[1-cos(x)]}\implies \cfrac{sin^2(x)-1+2cos(x)-cos^2(x)}{sin[1-cos(x)]} \\\\\\ \cfrac{sin^2(x)-[\underline{sin^2(x)+cos^2(x)}]+2cos(x)-cos^2(x)}{sin[1-cos(x)]} \\\\\\ \cfrac{sin^2(x)-sin^2(x)-cos^2(x)+2cos(x)-cos^2(x)}{sin[1-cos(x)]}$

$\bf \cfrac{-2cos^2(x)+2cos(x)}{sin[1-cos(x)]}\implies \cfrac{2cos(x)[-cos(x)+1]}{sin[1-cos(x)]}\implies \cfrac{2cos(x)[\underline{1-cos(x)}]}{sin[\underline{1-cos(x)}]} \\\\\\ \cfrac{2cos(x)}{sin(x)}\implies 2\cdot \cfrac{cos(x)}{sin(x)}\implies 2cot(x)$