Question 66295
tan^2x - sin^2x - tan ^2xsin^2x
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Then converting the tan^2 to sin^2/cos^2 you get:

(sin^2/cos^2)-sin^2 - (sin^2/cos^2)(sin^2)

Factor out the sin^2 to get:
(sin^2)[1/cos^2 - 1 -(sin^2/cos)^2]
Rewrite the 2nd factor with LCD=cos^2 to get:
= (sin^2)[1-cos^2-sin^2]/cos^2
= (sin^2/cos^2)[1-(cos^2+sin^2)]
But cos^2+sin^2=1, SO you get,
=(sin^2/cos^2)[1-1]
=0
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Cheers,
Stan H.