Yet another identity problem - (sin^2 B - tan^2 B) / 1 - sec^2 B = sin ^2 B

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Yet another identity problem - (sin^2 B - tan^2 B) / 1 - sec^2 B = sin ^2 B

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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We know sin^2 = 2 sin cos
1 - sec^2 = tan^2
-(sin^2 - tan^2) -------------- = sin^2 tan^2

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Other answers:

-sin^2 tan^2 ----- + ----- = sin^2 tan^2 tan^2
-sin^2 --------- + 1 = sin^2 sin^2 ---- cos^2
flip the bottom and multiply to get: -cos^2 +1 = sin^2 1 - cos^2 = sin^2 sin^2 = sin^2
Oops, the original equation's numerator is (sin^2 - tan^2). I'm so sorry for typing that wrong! Thanks for doing it, it must have been hard.
........... and I got outta bed for this? :)
multiply both sides by (-1) and you get the same result.... I think
Yes, you do, and thank you soooooo much and sorry for the mistake!

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