Trigonometry


  1. The value of
    4
    =
    3
    + 3 sin² α will be
    1 + cot²α1 + cot²α









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    4
    +
    1
    + 3 sin²α
    1 + tan²α1 + cot²α

    4
    +
    1
    + 3 sin²α
    sec²αcosec²α

    = 4 cos²α + sin²α + 3 sin²α
    = 4(cos²α + sin²α) = 4

    Correct Option: A

    4
    +
    1
    + 3 sin²α
    1 + tan²α1 + cot²α

    4
    +
    1
    + 3 sin²α
    sec²αcosec²α

    = 4 cos²α + sin²α + 3 sin²α
    = 4(cos²α + sin²α) = 4


  1. The numerical value of
    1
    =
    3
    + 2 sin² θ will be
    1 + cot²θ1 + tan²θ









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    1
    +
    3
    + 2sin² θ
    1 + cot² θ1 + tan² θ

    =
    1
    +
    3
    + 2sin² θ
    cosec² θsec² θ

    = sin² θ + 3cos² θ + 2sin² θ
    = 3(sin² θ + cos² θ) = 3

    Correct Option: D

    1
    +
    3
    + 2sin² θ
    1 + cot² θ1 + tan² θ

    =
    1
    +
    3
    + 2sin² θ
    cosec² θsec² θ

    = sin² θ + 3cos² θ + 2sin² θ
    = 3(sin² θ + cos² θ) = 3



  1. If x cos θ – y sin θ = √x² + y²
    cos²θ
    +
    sin²θ
    =
    1
    , then the correct relation is
    x² + y²









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    According to question,
    x cosθ – y sinθ = √x² + y²...(i)

    cos²θ
    +
    sin²θ
    =
    1
    ...(ii)
    x² + y²


    sinθ =
    - y
    x² + y²

    cosθ =
    x
    x² + y²

    From equation (i)
    x
    cosθ -
    y
    sinθ = 1
    x² + y²x² + y²

    cos²θ
    +
    sin²θ
    =
    1
    x² + y²

    +
    =
    1
    (x² + y²)a²(x² + y²)b²x² + y²

    +
    = 1

    Correct Option: B

    According to question,
    x cosθ – y sinθ = √x² + y²...(i)

    cos²θ
    +
    sin²θ
    =
    1
    ...(ii)
    x² + y²


    sinθ =
    - y
    x² + y²

    cosθ =
    x
    x² + y²

    From equation (i)
    x
    cosθ -
    y
    sinθ = 1
    x² + y²x² + y²

    cos²θ
    +
    sin²θ
    =
    1
    x² + y²

    +
    =
    1
    (x² + y²)a²(x² + y²)b²x² + y²

    +
    = 1


  1. If cos x + cos²x = 1, the numerical value of (sin12x + 3 sin10x + 3 sin8x + sin6x – l) is :









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    cos x + cos² x = 1
    ⇒ cos x = 1 – cos² x = sin² x ...(i)
    ∴ sin12x + 3 sin10x + 3 sin8x + sin6x – 1
    = (sin4 x + sin2 x)3 – 1
    = (cos2x + sin2x)3 – 1 [By (i)]
    = 1 – 1 = 0

    Correct Option: C

    cos x + cos² x = 1
    ⇒ cos x = 1 – cos² x = sin² x ...(i)
    ∴ sin12x + 3 sin10x + 3 sin8x + sin6x – 1
    = (sin4 x + sin2 x)3 – 1
    = (cos2x + sin2x)3 – 1 [By (i)]
    = 1 – 1 = 0



  1. If
    sec θ + tan θ
    =
    5
    , then sin θ is equal to
    sec θ - tan θ3









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    secθ + tanθ
    =
    5
    secθ - tanθ3

    ⇒ 5 sec θ – 5 tan θ
    = 3 sec θ + 3 tan θ
    ⇒ 2 sec θ = 8 tan θ
    =
    tanθ
    =
    2
    =
    1

    secθ84

    sinθ
    × cosθ =
    1
    cosθ4

    ⇒ sinθ =
    1
    4

    Correct Option: A

    secθ + tanθ
    =
    5
    secθ - tanθ3

    ⇒ 5 sec θ – 5 tan θ
    = 3 sec θ + 3 tan θ
    ⇒ 2 sec θ = 8 tan θ
    =
    tanθ
    =
    2
    =
    1

    secθ84

    sinθ
    × cosθ =
    1
    cosθ4

    ⇒ sinθ =
    1
    4