Question

1-sin^2x/sinx cosx=ctgx-tgx

Answer 1

$\frac{1-sin^{2}x}{sinxcosx} = ctgx-tgx$; $\frac{ cos^{2}x }{sinxcosx} =ctgx-tgx$; $ctgx=ctgx-tgx$ ; $tgx=0$ ;
$x= \pi n, n$∈ Z