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anonymous

  • one year ago

-tan^2x + sec^2x=1

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  1. mathstudent55
    • one year ago
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    Do you need to prove this identity?

  2. anonymous
    • one year ago
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    Yes

  3. mathstudent55
    • one year ago
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    Rewrite tan x and sec x in terms of sin x and cos x. Then multiply both sides by \(\cos^2 x\)

  4. anonymous
    • one year ago
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    Sec^2x=1/cos^2x

  5. mathstudent55
    • one year ago
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    yes. what about tan? what is tan in terms of sin and cos?

  6. anonymous
    • one year ago
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    I'm not sure😓

  7. anonymous
    • one year ago
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    NO

  8. anonymous
    • one year ago
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    U GAYY

  9. triciaal
    • one year ago
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    |dw:1437335013169:dw|

  10. triciaal
    • one year ago
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    |dw:1437335106407:dw|

  11. mathstudent55
    • one year ago
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    Right. tan x = sin x / cos x

  12. mathstudent55
    • one year ago
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    \(-\tan^2x + \sec^2x=1\) \(-\dfrac{\sin^2x}{\cos^2 x} + \dfrac{1}{\cos^2x}=1\) Now multiply both sides by \(\cos^2 x\)

  13. anonymous
    • one year ago
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    ok :)

  14. anonymous
    • one year ago
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    -sin^2x/cos^2x=cos^2x?

  15. mathstudent55
    • one year ago
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    No

  16. mathstudent55
    • one year ago
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    \(-\dfrac{\sin^2x}{\cos^2 x} + \dfrac{1}{\cos^2x}=1\) Now we multiply both sides by \(\cos ^2 x\) \( \cos^2 x(-\dfrac{\sin^2x}{\cos^2 x} + \dfrac{1}{\cos^2x})=1 \times \cos^2 x\) \( -\dfrac{\sin^2x \cancel{\cos^2 x}}{\cancel{\cos^2 x}~~1} + \dfrac{\cancel{\cos^2 x}~~1}{\cancel{\cos^2x}~~1})= \cos^2 x\) \(- \sin^2 x + 1 = \cos^2 x\) \(1 = \sin^2 x + \cos^2 x\) \(\sin^2 x + \cos^2 x = 1\) The last equation is a well-known trig identity.

  17. anonymous
    • one year ago
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    thank you >.< ill work harder to memorize the identities

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