Question

(1+cosx) (cscx-cotx) i need to simplify it

Answer 1

\(\bf [1-cos(x)][csc(x)-cot(x)]\implies [1-cos(x)]\left[\cfrac{1}{sin(x)}-\cfrac{cos(x)}{sin(x)}\right]\\ \quad \\\
[1-cos(x)]\left[\cfrac{1-cos(x)}{sin(x)}\right]\implies \cfrac{[1-cos(x)][1-cos(x)]}{sin(x)}\\ \quad \\
\textit{recall that}\quad (a-b)(a+b) = a^2-b^2\qquad thus\\ \quad \\
\implies \cfrac{1^2-cos^2(x)}{sin(x)}\)

now can you recall what \(\bf 1^2-cos^2(x)\) equals to ? check the identity of \(\bf sin^2(\theta)+cos^2(\theta)=1\)

Answer 2

hmmm is supposed to be 1+cos(x) so \(\bf [1+cos(x)][csc(x)-cot(x)]\implies [1+cos(x)]\left[\cfrac{1}{sin(x)}-\cfrac{cos(x)}{sin(x)}\right]\\ \quad \\\
[1+cos(x)]\left[\cfrac{1-cos(x)}{sin(x)}\right]\implies \cfrac{[1+cos(x)][1-cos(x)]}{sin(x)}\\ \quad \\
\textit{recall that}\quad (a-b)(a+b) = a^2-b^2\qquad thus\\ \quad \\
\implies \cfrac{1^2-cos^2(x)}{sin(x)}\)

now can you recall what \(\bf 1^2-cos^2(x)\) equals to ? check the identity of \(\bf sin^2(\theta)+cos^2(\theta)=1\)