Question

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

Answer 1

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

and surely you know what that is