Question

Tanx-cotx / sinxcosx =sec^2-csc^2x. Please show all steps.

Answer 1

`\bf \cfrac{tan(x)-cot(x)}{sin(x)cos(x)}\implies \cfrac{(sin(x))/(cos(x))-(cos(x))/(sin(x))}{sin(x)cos(x)}\implies \cfrac{(sin^2(x)-cos^2(x))/(cos(x)sin(x))}{(sin(x)cos(x))/(1)} \\\\\\ \cfrac{sin^2(x)-cos^2(x)}{cos(x)sin(x)}\cdot \cfrac{1}{sin(x)cos(x)}\implies \cfrac{sin^2(x)-cos^2(x)}{cos^2(x)sin^2(x)} \\\\\\ \textit{and now, we distribute the denominator} \\\\\\ \cfrac{sin^2(x)}{cos^2(x)sin^2(x)}-\cfrac{cos^2(x)}{cos^2(x)sin^2(x)}\implies \cfrac{1}{cos^2(x)}-\cfrac{1}{sin^2(x)}`

and surely you know what that is