0944399
contestada

Write each expression in terms of sine and​ cosine, and then simplify so that no quotients appear in the final expression and all functions are of x only.
[tex]\frac{csc^2 (-x)-1}{1-cos^2 (-x)}[/tex]

Respuesta :

[tex]\stackrel{\textit{reciprocal ident.}}{csc(\theta)=\cfrac{1}{sin(\theta)}} ~\hfill \stackrel{\textit{symmetry ident.}}{sin(-\theta )=-sin(\theta)}~\hfill \begin{array}{llll} \stackrel{\textit{pythagorean ident.}}{sin^2(\theta)+cos^2(\theta)=1}\\ sin^2(\theta)=1-cos^2(\theta)\\ cos^2(\theta)=1-sin^2(\theta) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]

[tex]\cfrac{csc^2(-x)-1}{1-cos^2(-x)}\implies \cfrac{~~\frac{1}{sin^2(-x)}-1~~}{sin^2(-x)}\implies \cfrac{~~ \frac{1-sin^2(-x)}{sin^2(-x)}~~}{\frac{sin^2(-x)}{1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{keeping in mind that}~\hfill }{sin^2(-x)\implies [sin(-x)]^2\implies [-sin(x)]^2\implies [sin(x)]^2\implies sin^2(x)} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\cfrac{~~ \frac{1-sin^2(x)}{sin^2(x)}~~}{\frac{sin^2(x)}{1}}\implies \cfrac{1-sin^2(x)}{sin^2(x)}\cdot \cfrac{1}{sin^2(x)}\implies \cfrac{cos^2(x)}{sin^2(x)}\cdot \cfrac{1}{sin^2(x)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill cot^2(x)csc^2(x)~\hfill[/tex]