Question

(cos^2t - ctg^2t)/(sin^2t - tg^2t)

Answer 1

$\frac{Cos^{2}t-Ctg^{2}t }{Sin{2}t-tg^{2}t}=\frac{Cos^{2}t-\frac{Cos^{2}t }{Sin^{2}t}}{Sin^{2}t-\frac{Sin^{2}t }{Cos^{2}t }} =\frac{\frac{Sin^{2}tCos^{2}t-Cos^{2}t}{Sin^{2}t}}{\frac{Sin^{2}tCos^{2}t-Sin^{2}t}{Cos^{2}t }}=\frac{Cos^{2}t(Sin^{2}t-1)*Cos^{2}t }{Sin^{2}t(Cos^{2}t-1)*Sin^{2}t}=\frac{-Cos^{2}t*Cos^{4}t}{-Sin^{2}t*Sin^{4}t}=\frac{Cos^{6}t }{Sin^{6}t }=Ctg^{6}t$