Question

I just need help with the first step for verifying this trigonometric identity, which is to get the common denominator.

Answer 1

`\textit{Pythagorean Identities} \\\\ 1+\cot^2(\theta)=\csc^2(\theta)\implies \cot^2(\theta)=\csc^2(\theta)-1 \\\\[-0.35em] \rule{34em}{0.25pt}`

`\cfrac{1}{\csc(x)+1}+\cfrac{1}{\csc(x)-1}~~ = ~~2\sec(x)\tan(x) \\\\[-0.35em] ~\dotfill\\\\ \cfrac{1}{\csc(x)+1}+\cfrac{1}{\csc(x)-1}\implies \cfrac{[\csc(x)-1]~~ + ~~[\csc(x)+1]}{\underset{ \textit{difference of squares} }{[\csc(x)+1][\csc(x)-1]}} \\\\\\ \cfrac{[\csc(x)-1]~~ + ~~[\csc(x)+1]}{\csc^2(x)-1^2}\implies \cfrac{2\csc(x)}{\csc^2(x)-1}\implies \cfrac{2\csc(x)}{\cot^2(x)}`

`2\cdot \cfrac{\csc(x)}{\cot(x)\cot(x)}\implies 2\cdot \cfrac{\csc(x)}{\cot(x)}\cdot \cfrac{1}{\cot(x)}\implies 2\cdot \cfrac{\csc(x)}{\cot(x)}\cdot \tan(x) \\\\\\ 2\cdot \cfrac{~~ ( 1 )/( \sin(x) ) ~~}{(\cos(x))/(\sin(x))}\cdot \tan(x)\implies 2\cdot \cfrac{ 1 }{ \sin(x) }\cdot \cfrac{\sin(x)}{\cos(x)}\cdot \tan(x) \\\\\\ 2\cdot \cfrac{1}{\cos(x)}\cdot \tan(x)\implies 2\sec(x)\tan(x)`