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gjhfdfg

  • 3 years ago

Verify the identity. cot ( theta - pi / 2 ) = - tan theta

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  1. Zekarias
    • 3 years ago
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    \[\cot(x-\pi/2)=\tan(\pi/2-(x-\pi/2))=\tan(-x)=-\tan(x)\]

  2. Kashan
    • 3 years ago
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    cot ( theta - pi / 2 )= cos ( theta - pi / 2 ) / sin ( theta - pi / 2 ) cos ( theta - pi / 2 )= cos(theta) cos (pi/2) + sin(theta) sin (pi/2) = cos(theta) (0) + sin(theta) (1) = sin(theta) sin ( theta - pi / 2 )= sin(theta) cos(pi/2) - cos(theta) sin(pi/2) = sin(theta) (0) - cos(theta) (1) = -cos(theta) so you get, cot ( theta - pi / 2 )= sin(theta) / -cos(theta) cot ( theta - pi / 2 )= -tan(theta)

  3. Zekarias
    • 3 years ago
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    For any CO - trigonometric functions m(x) and co-m(x), this property is always true.\[m(x)=co -m(\pi/2-x)=co -m(x-\pi/2).....or ..vise..versa----if ..m(x)..is ..even\]\[m(x)=co -m(\pi/2-x)=-co -m(x-\pi/2).....or ..vise..versa----if ..m(x)..is ..odd\]

  4. Kashan
    • 3 years ago
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    got it now..??

  5. Zekarias
    • 3 years ago
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    The first one is for even function and the second one for odd.

  6. Kashan
    • 3 years ago
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    @gjhfdfg ????

  7. gjhfdfg
    • 3 years ago
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    I think so.?

  8. gjhfdfg
    • 3 years ago
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    Alright I get it thank you.!

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