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cos(270^@-theta) =...

`cos(270^@-theta) =`

A

`-cos theta`

B

`-sin theta`

C

`sin theta`

D

`cos theta`

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Knowledge Check

  • If 4 cos theta.cos(120^(@)+theta).cos(120^(@)-theta)=1//2 then theta =

    A
    `n pi` or `(3n pm 1) (pi)/(9) : n in Z`
    B
    `(n pi)/(3)+(-1)^(n)(pi)/(6) : n in Z`
    C
    `(2n pi)/(3)pm(pi)/(9) : n in Z`
    D
    `(n pi)/(3)+(pi)/(18) : n in Z`

  • If tan theta = -(5)/(12) and theta is not in the fourth quadrant then ( tan (90^(@)+ theta ) - sin(180^(@) - theta ))/( sin (270^(@)- theta ) + " cosec " (360^(@) - theta ))=

    A
    `(109)/(131)`
    B
    `-(109)/(131)`
    C
    `(131)/(109)`
    D
    `-(131)/(109)`

  • cos theta - cos (60^(@) + theta ) - cos(60^(@) - theta )=

    A
    0
    B
    1
    C
    `1//4`
    D
    `3//4`

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    cosec(270-theta) = ……………..

    Show that cos theta+cos(120^(@)+theta)+cos(240^(@)+theta)=0

    cos theta cos (60^(@) + theta ) * cos (60^(@) - theta ) =

    cos theta + cos (240^(@) + theta ) + cos (240^(@)- theta )=

    cos theta + cos (120^(@) + theta ) + cos (120^(@) - theta )=

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