Prove:
sin^2x (sec^2x + csc^2x)=sec^2x

Sin^2x (sec^2x + csc^2x) = sec^2x

I would convert the functions in the parentheses to their reciprocals.

sin^2x (1/cos^2x + 1/sin^2x) = sec^2x

Now distribute the sine.

sin^2x/cos^2x + sin^2x/sin^2x = sec^2x

Remember that sine divided by cosine is always tangent.

tan^2x + sin^2x/sin^2x = sec^2x

The remaining fraction is simply 1.

tan^2x + 1 = sec^2x

Use the Pythagorean identity to add the left side.

sec^2x = sec^2x

Q.E.D.

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