Solve the equationfrac{cot(x)-tan(x)}{(cot^{2}(x)-tan^{2}(x)}=sin xcos x

Apply Difference of Two Squares Formula:
$x^2-y^2=(x+y)(x-y)$
$\to {\cot }^2(x)-{\tan }^2(x)=(\cot (x)+\tan (x))(\cot (x)-\tan (x))$
$⟹\frac{\cot (x)-\tan (x)}{{\cot }^2(x)-{\tan }^2(x)}=\frac{\cot (x)-\tan (x)}{(\cot (x)+\tan (x))(\cot (x)-\tan (x))}=\frac{1}{(\cot (x)+\tan (x)}$
$=\frac{1}{2\csc (2x)}$
$=\frac{\sin (2x)}{2}$
$=\sin x\cos x$
Therefore
$\sqrt{a^2-a^2{\sin }^2(\theta )}=a\cos (\theta )$