Question

Prove (1/(1+cosx)) - (1/(1-cosx)

1               ;     1
------------- - ------------
(1 + cosx) (1 - cosx)

Answer 1

^^

Explanation:

Answer 2

`\bf \cfrac{1}{1+cos(x)}-\cfrac{1}{1-cos(x)} \\\\\\ \textit{let us take the LCD of }[1+cos(x)][1-cos(x)]\qquad thus \\\\\\ \cfrac{[1-cos(x)]-[1+cos(x)]}{[1+cos(x)][1-cos(x)]}\implies \cfrac{1-cos(x)-1-cos(x)}{[1+cos(x)][1-cos(x)]} \\\\\\\cfrac{-2cos(x)}{[1+cos(x)][1-cos(x)]}\\\\ -----------------------------\\\\ \textit{now recall your }\textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\`

`\bf -----------------------------\\\\ \cfrac{-2cos(x)}{[1+cos(x)][1-cos(x)]}\implies \cfrac{-2cos(x)}{1^2-cos^2(x)}\\\\ -----------------------------\\\\ \textit{now recall your pythagorean identities}\\\\ sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\\\\ -----------------------------\\\\ \cfrac{-2cos(x)}{1^2-cos^2(x)}\implies \cfrac{-2cos(x)}{sin^2(x)}\implies \cfrac{-2cos(x)}{sin(x)sin(x)} \\\\\\ -2\cfrac{cos(x)}{sin(x)}\cdot \cfrac{1}{sin(x)}\implies -2cot(x)csc(x)`