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sin ^(2) " (pi)/(6) + cos ^(2) "" (pi)/(...

`sin ^(2) " (pi)/(6) + cos ^(2) "" (pi)/(3) - tan ^(2) " (pi)/(4) =- 1/2`

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Step by step text solution for sin ^(2) " (pi)/(6) + cos ^(2) "" (pi)/(3) - tan ^(2) " (pi)/(4) =- 1/2 by MATHS experts to help you in doubts & scoring excellent marks in Class 11 exams.

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2 sin ^(2) "" (pi)/(6) + cosec ^(2) "" (7pi)/(6) cos ^(2) "" (pi)/(3) = 3/2

2 sin ^(2) "" (3pi)/(4) + 2 cos ^(2) "" (pi)/( 4) + 2 sec ^(2) "" (pi)/(3) =10

Knowledge Check

  • cos^(2)""(3pi)/(5)+cos^(2)""(4pi)/(5)=

    A
    `4//5`
    B
    `5//2`
    C
    `5//4`
    D
    `3//4`

  • cos^(2)""(pi)/(5)+ sin^(2)""(4pi)/(5)=

    A
    0
    B
    `1//2`
    C
    `1//5`
    D
    1

  • sin^(2)""(2pi)/(3)+ cos^(2)""(5pi)/6-tan^(2)""(3pi)/4=

    A
    `0`
    B
    `1//2`
    C
    `1//5`
    D
    1

    • Similar Questions

    Explore conceptually related problems

    cot ^(2) "" (pi)/(6) + cosec ""(5pi)/(6) + 3 tan ^(2) ""(pi)/(6) =6

    Evaluate (ii) sin^(2). (2pi)/(3) + cos^(2). (5pi)/(6) - tan^(2). (3 pi)/(4)

    cos^(2)""((pi)/(4) +x ) - sin^(2) ""((pi)/4- x ) =

    To find the sum "sin"^(2) (2pi)/(7) + "sin"^(2) (4pi)/(7) + "sin"^(2) (8pi)/(7) , we follow the following method. Put 7 theta = 2 n pi , where n is any integer. Then sin 4 theta = sin (2n pi - 3 theta) = - sin 3 theta" "…(i) This means that sin theta takes the values 0. +- sin (2pi//7), +- sin (4pi//7), and +- sin (8pi//7) . From Eq. (i), we now get 2 sin 2theta cos 2 theta = 4 sin^(3) theta - 3 sin theta or 4 sin theta cos theta(1 - 2 sin^(2) theta) = (4 sin^(2) theta - 3) sin theta Rejecting the value sin theta = 0 , we get 4 cos theta (1-2 sin^(2) theta) = 4 sin^(2) theta - 3 or 16 cos^(2) theta (1-2sin^(2)theta)^(2) = (4 sin^(2) theta - 3)^(2) or 16(1-sin^(2)theta)(1-4 sin^(2)theta + 4 sin^(4) theta) = 16 sin^(4) theta - 24 sin^(2) theta + 9 or 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta - 7 = 0 , and this is cubic in sin^(2) theta with the roots sin^(2) ((2pi)/(7)),sin^(2)((4pi)/(7))and sin^(2) ((8pi)/(7)) The sum of these roots is "sin"^(2) (2pi)/(7) + "sin"^(2) (4pi)/(7) + "sin"^(2) (8pi)/(7) = (112)/(64) = (7)/(4) . The value of ("tan"^(2) (pi)/(7)+"tan"^(2)(2pi)/(7)+"tan"^(2)(3pi)/(7))xx("cot"^(2)(pi)/(7)+"cot"^(2) (2pi)/(7)+"cot"^(2) (3pi)/(7)) is

    To find the sum "sin"^(2) (2pi)/(7) + "sin"^(2) (4pi)/(7) + "sin"^(2) (8pi)/(7) , we follow the following method. Put 7 theta = 2 n pi , where n is any integer. Then sin 4 theta = sin (2n pi - 3 theta) = - sin 3 theta" "…(i) This means that sin theta takes the values 0. +- sin (2pi//7), +- sin (4pi//7), and +- sin (8pi//7) . From Eq. (i), we now get 2 sin 2theta cos 2 theta = 4 sin^(3) theta - 3 sin theta or 4 sin theta cos theta(1 - 2 sin^(2) theta) = (4 sin^(2) theta - 3) sin theta Rejecting the value sin theta = 0 , we get 4 cos theta (1-2 sin^(2) theta) = 4 sin^(2) theta - 3 or 16 cos^(2) theta (1-2sin^(2)theta)^(2) = (4 sin^(2) theta - 3)^(2) or 16(1-sin^(2)theta)(1-4 sin^(2)theta + 4 sin^(4) theta) = 16 sin^(4) theta - 24 sin^(2) theta + 9 or 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta - 7 = 0 , and this is cubic in sin^(2) theta with the roots sin^(2) ((2pi)/(7)),sin^(2)((4pi)/(7))and sin^(2) ((8pi)/(7)) The sum of these roots is "sin"^(2) (2pi)/(7) + "sin"^(2) (4pi)/(7) + "sin"^(2) (8pi)/(7) = (112)/(64) = (7)/(4) . The value of ("tan"^(2)(pi)/(7)+"tan"^(2)(2pi)/(7)+"tan"^(2) (3pi)/(7))/("cot"^(2)(pi)/(7)+"cot"^(2) (2pi)/(7) + "cot"^(2) (3pi)/(7)) is

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