Trigonometry


  1. If θ be positive acute angle and 5 cosθ + 12 sinθ = 13, then the value of cosθ is









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    5 cosθ + 12 sinθ = 13
    ⇒  5 cosθ – 13 = 12 sinθ
    On squaring both sides,
    25 cos2θ + 169 – 130 cosθ
    = 144 (1 – cos2θ)
    ⇒  cos2θ – 130 cosθ + 169
    = 144 – 144 cos2θ
    ⇒  144 cos2θ + 25 cos2θ – 130 cosθ + 169 – 144 = 0
    ⇒  169 cos2θ – 130 cosθ + 25 = 0
    ⇒  (13 cosθ – 5)2 = 0
    ⇒  13 cosθ – 5 = 0
    ⇒  13 cosθ = 5

    ⇒  cosθ =
    5
    13

    OR
    5 cosθ + 12 sinθ = 13
    ⇒ 
    5
    cosθ +
    12
    sinθ = 1
    1313

    ∵  cos2θ + sin2θ = 1
    ∴  cosθ =
    5
    13

    Correct Option: B

    5 cosθ + 12 sinθ = 13
    ⇒  5 cosθ – 13 = 12 sinθ
    On squaring both sides,
    25 cos2θ + 169 – 130 cosθ
    = 144 (1 – cos2θ)
    ⇒  cos2θ – 130 cosθ + 169
    = 144 – 144 cos2θ
    ⇒  144 cos2θ + 25 cos2θ – 130 cosθ + 169 – 144 = 0
    ⇒  169 cos2θ – 130 cosθ + 25 = 0
    ⇒  (13 cosθ – 5)2 = 0
    ⇒  13 cosθ – 5 = 0
    ⇒  13 cosθ = 5

    ⇒  cosθ =
    5
    13

    OR
    5 cosθ + 12 sinθ = 13
    ⇒ 
    5
    cosθ +
    12
    sinθ = 1
    1313

    ∵  cos2θ + sin2θ = 1
    ∴  cosθ =
    5
    13


  1. If tanθ + cotθ = 5, then the value of tan2θ + cot2θ is









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    tanθ + cotθ = 5
    On squaring both sides,
    (tanθ + cotθ)2 = 25
    ⇒  tan2θ + cot2θ + 2tanθ. cotθ = 25
    ⇒  tan2θ + cot2θ + 2 = 25
    ⇒  tan2θ + cot2θ = 25 – 2 = 23

    Correct Option: C

    tanθ + cotθ = 5
    On squaring both sides,
    (tanθ + cotθ)2 = 25
    ⇒  tan2θ + cot2θ + 2tanθ. cotθ = 25
    ⇒  tan2θ + cot2θ + 2 = 25
    ⇒  tan2θ + cot2θ = 25 – 2 = 23



  1. If 7sin2θ + 3cos2θ = 4 then the value of secθ + cosecθ is









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    7sin2θ + 3cos2θ = 4
    ⇒  4 sin2θ + 3 sin2θ + 3 cos2θ = 4
    ⇒  4 sin2θ + 3(sin2θ + cos2θ) = 4
    ⇒  4 sin2θ = 4 – 3
    [∵  sin2θ + cos2θ = 1]
    ⇒  4 sin2θ = 1

    ⇒  sin2θ =
    1
    4

    ⇒  sinθ =
    1
    = sin 30°
    2

    ⇒  θ = 30°
    ∴  secθ + cosecθ
    = sec 30° + cosec 30°
    =
    2
    + 2
    3

    Correct Option: B

    7sin2θ + 3cos2θ = 4
    ⇒  4 sin2θ + 3 sin2θ + 3 cos2θ = 4
    ⇒  4 sin2θ + 3(sin2θ + cos2θ) = 4
    ⇒  4 sin2θ = 4 – 3
    [∵  sin2θ + cos2θ = 1]
    ⇒  4 sin2θ = 1

    ⇒  sin2θ =
    1
    4

    ⇒  sinθ =
    1
    = sin 30°
    2

    ⇒  θ = 30°
    ∴  secθ + cosecθ
    = sec 30° + cosec 30°
    =
    2
    + 2
    3


  1. If x2 = sin230° + 4 cot245° – sec260°, then the value of x (x > 0) is









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    x2 = sin230° + 4 cot245° – sec260°

    =
    1
    2+ 4(1)2 – (2)2
    2

    =
    1
    + 4 – 4 =
    1
    44

    ∴  x =
    1
    2

    Correct Option: D

    x2 = sin230° + 4 cot245° – sec260°

    =
    1
    2+ 4(1)2 – (2)2
    2

    =
    1
    + 4 – 4 =
    1
    44

    ∴  x =
    1
    2



  1. If
    cos θ
    +
    cos θ
    = 4 , then the value of θ (0° < θ < 90 ° ) is
    1 - sin θ1 + sin θ










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    cos θ
    +
    cos θ
    = 4
    1 - sin θ1 + sin θ

    cos θ(1 + sin θ) + cos θ(1 - sin θ)
    = 4
    (1 - sin θ)(1 + sin θ)

    cos θ + cos θ.sin θ + cos θ - cos θ.sin θ
    = 4
    (1 - sin2 θ)

    2cos θ
    = 4
    cos2 θ

    1
    = 2
    cos θ

    ⇒ cos θ =
    1
    = cos 60° ⇒ θ = 60°
    2

    Correct Option: A

    cos θ
    +
    cos θ
    = 4
    1 - sin θ1 + sin θ

    cos θ(1 + sin θ) + cos θ(1 - sin θ)
    = 4
    (1 - sin θ)(1 + sin θ)

    cos θ + cos θ.sin θ + cos θ - cos θ.sin θ
    = 4
    (1 - sin2 θ)

    2cos θ
    = 4
    cos2 θ

    1
    = 2
    cos θ

    ⇒ cos θ =
    1
    = cos 60° ⇒ θ = 60°
    2