Question

Identidades trigonometricas

Answer 1

${ \dfrac{\cos^2x+\tan^2x-1}{\sin^2x} =}\\\\\\{ \dfrac{\cos^2x+\frac{\sin^2x}{\cos^2x}-1}{\sin^2x} =}\\\\\\{ \dfrac{ \frac{\cos^4x+\sin^2x-\cos^2x}{\cos^2x} }{\sin^2x} =}\\\\\\{ \dfrac{ \frac{\cos^4x+1-\cos^2x-\cos^2x}{\cos^2x} }{\sin^2x} =}\\\\\\{ \frac{ \dfrac{\cos^4x-2\cos^2x+1}{\cos^2x} }{\sin^2x} =}\\\\\\{ \dfrac{ \frac{(1-\cos^2x)^2}{\cos^2x} }{\sin^2x} =}\\\\\\{ \dfrac{(\sin^2x)^2}{\cos^2x} \times \dfrac{1}{\sin^2x} =}\\\\\\{ \dfrac{\sin^2x}{\cos^2x} =}\\\\\\{\boxed{\tan^2x}}$

Salu2.!! :)
Wellington