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Mathematics 10 Online
OpenStudy (anonymous):

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

OpenStudy (amistre64):

We know sin^2 = 2 sin cos

OpenStudy (amistre64):

1 - sec^2 = tan^2

OpenStudy (amistre64):

-(sin^2 - tan^2) -------------- = sin^2 tan^2

OpenStudy (amistre64):

-sin^2 tan^2 ----- + ----- = sin^2 tan^2 tan^2

OpenStudy (amistre64):

-sin^2 --------- + 1 = sin^2 sin^2 ---- cos^2

OpenStudy (amistre64):

flip the bottom and multiply to get: -cos^2 +1 = sin^2 1 - cos^2 = sin^2 sin^2 = sin^2

OpenStudy (anonymous):

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.

OpenStudy (amistre64):

........... and I got outta bed for this? :)

OpenStudy (amistre64):

multiply both sides by (-1) and you get the same result.... I think

OpenStudy (anonymous):

Yes, you do, and thank you soooooo much and sorry for the mistake!

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