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haleyking345

  • 3 years ago

tan^2x - tan^2x sin^2x = sin^2x How in the world could you solve this?

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  1. TuringTest
    • 3 years ago
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    \[\tan^2x-\tan^2x\sin^2x=\tan^2x(1-\sin^2x)\]and what is the identity for \(1-\sin^2x\) ?

  2. haleyking345
    • 3 years ago
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    cos^2x

  3. TuringTest
    • 3 years ago
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    right, so now what do you have on the left?

  4. haleyking345
    • 3 years ago
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    tan^2x-tan^2xsin^2x=tan^2x

  5. TuringTest
    • 3 years ago
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    no, I mean after what we just did...

  6. haleyking345
    • 3 years ago
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    tan^2x

  7. TuringTest
    • 3 years ago
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    \[\tan^2x-\tan^2x\sin^2x=\tan^2x(1-\sin^2x)=\tan^2x\cos^2x=?\]

  8. TuringTest
    • 3 years ago
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    use\[\tan x=\frac{\sin x}{\cos x}\]

  9. haleyking345
    • 3 years ago
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    that would leave you with sin^2x as the answer

  10. TuringTest
    • 3 years ago
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    which is what you have on the left, which proves the identity if you want a solution, the answer is \(\mathbb R\), but I think the point here was just to prove the identity.

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