Respuesta :
Given:
[tex](cot^2x+1)(sin^2x-1)=-cotx[/tex]Let's verify the identity.
Apply Pythagorean identity on the left:
[tex]\begin{gathered} (1+cos^2x)(-1+sin^2x)=-cotx \\ \\ csc^2x(-cos^2x)=-cotx \end{gathered}[/tex]The next step is to apply reciprocal identity.
[tex]\begin{gathered} \text{ Where:} \\ csc^2x=(\frac{1}{sinx})^2 \end{gathered}[/tex]Thus, we have:
[tex]\begin{gathered} (\frac{1}{sinx})^2(-cos^2x)=-cotx \\ \\ \frac{1^2}{sin^2x}(-cos^2x)=-cotx \end{gathered}[/tex]Solving further:
[tex]\frac{-cos^2x}{sin^2x}=-cotx[/tex]Apply quotient identity:
[tex]\begin{gathered} \text{ Where:} \\ cotx=\frac{cosx}{sinx} \end{gathered}[/tex]Using the quotient identity, we have:
[tex]-\frac{cos^2x}{sin^2x}=-cot^2x[/tex]Therefore, the given equation is an identity.
We have the following:
[tex]\begin{gathered} csc^2x(-cos^2x)====>\text{ Pythagorean Identity} \\ \\ \\ \frac{1^{2}}{s\imaginaryI n^{2}x}(-cos^2x)\text{ }====>\text{ Reciprocal identity} \\ \\ \\ -\frac{cos^2x}{s\imaginaryI n^2x}\text{ }===>\text{ Algebra} \\ \\ -cot^2x\text{ }====>\text{ Quotient identity} \end{gathered}[/tex]