SOLUTION: The "x's" are actually thetas but for simplicity sake I am going to use "x" instead tan^2x - sin^2x - tan ^2xsin^2x I tried to replace the tan^2x's with (sin^2x)/(cos^2x) but

Question 66295: The "x's" are actually thetas but for simplicity sake I am going to use "x" instead
tan^2x - sin^2x - tan ^2xsin^2x
I tried to replace the tan^2x's with (sin^2x)/(cos^2x) but I didn't know where to go from there.. or if that even works.
thanks alot.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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.
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