Question

$\frac{1 + tan ^{2}x }{csc x^{2} x} = sec x^{2} x - 1$

Answer 1

saludos 

recordar:

identidad pitagorica

1+ tan²x= sec²x

Answer 2

${\frac{1+tan^2x}{csc^2x}=}\\\\{\frac{1+\frac{sen^2}{cos^2x}}{\frac{1}{sen^2x}}=}\\\\{\frac{\frac{cos^2x+sen^2x}{cos^2x}}{\frac{1}{sen^2x}}=}\\\\{\frac{\frac{1}{cos^2x}}{\frac{1}{sen^2x}}=}\\\\{\frac{1}{cos^2x}\times \frac{sen^2x}{1}=}\\\\{\frac{sen^2x}{cos^2x}=}\\\\{tan^2x=}\\\\{\boxed{sec^2x-1}}$