we have the expression
[tex]\frac{secxcot^2x-cosx}{sin(-x)cot^2x}[/tex]
Rewrite the given expression
Remember that
sin(-x)=-sin(x)
[tex]\frac{\frac{1}{cosx}\frac{cos^2x}{sin^2x}-cosx}{-sinx\frac{cos^{2}x}{s\imaginaryI n^{2}x}}[/tex]
Simplify the expression
[tex]\begin{gathered} \frac{\frac{cosx}{s\imaginaryI n^2x}-cosx}{-\frac{cos^2x}{s\imaginaryI nx}} \\ \\ \frac{\frac{cosx-sin^2xcosx}{sin^2x}}{-\frac{cos^2x}{sinx}} \\ \\ \frac{cosx-s\imaginaryI n^{2}xcosx}{s\imaginaryI n^{2}x}\colon-\frac{cos^{2}x}{s\imaginaryI nx} \\ \\ \frac{sinx(cosx-sin^2xcosx)}{sin^2x(cos^2x)} \\ \\ \frac{(cosx-sin^2xcosx)}{sin^x(cos^2x)} \\ \\ \frac{cosx(1-s\imaginaryI n^2)}{s\imaginaryI nx(cos^2x)} \\ \\ \frac{(1-s\imaginaryI n^2)}{s\imaginaryI nx(cosx)} \\ \\ \frac{cos^2x}{s\imaginaryI nx(cosx)} \\ \\ \frac{cosx}{sinx} \\ \\ cotx \end{gathered}[/tex]
therefore
The answer is
yes, the expression is equivalent to cot(x)
