Question

I just confused myself trying this :

1/sin^2(x) + Sec^2(x)/Tan^2(x)

Answer 1

change everything in terms of sines and cosines

\[\sec^2 x = \frac{1}{\cos^2 x}\] \[\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\]

Answer 2

\[\large \frac{\sec^2 x}{\tan ^2 x} = \frac{\frac{1}{\cos^2 x}}{\frac{\sin^2 x}{\cos^2 x}}\]

does that help?