Trigonometry


  1. If sec θ + tan θ = 2 + √5 , then the value of sin θ + cos θ is :









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    sec θ + tan θ = 2 + √3

    ∴ sin θ - tan θ =
    1
    5 + 2

    [∵ sec² θ - tan² θ = 1]
    =
    5 - 2
    = √5 - 2
    (√5 + 2)(√5 - 2)

    On adding,
    2secθ = 2 + √5 + √5 - 2 = 2√5
    ⇒ secθ = √5 ⇒ cosθ =
    1
    5

    On subtracting,
    2tanθ= 2 + √5 - √5 + 2 = 4
    ⇒ tan θ = 2
    tanθ
    = sin θ
    2
    sec θ5

    ∴ sin θ + cos θ
    2
    +
    1
    55

    =
    3
    5

    Correct Option: A

    sec θ + tan θ = 2 + √3

    ∴ sin θ - tan θ =
    1
    5 + 2

    [∵ sec² θ - tan² θ = 1]
    =
    5 - 2
    = √5 - 2
    (√5 + 2)(√5 - 2)

    On adding,
    2secθ = 2 + √5 + √5 - 2 = 2√5
    ⇒ secθ = √5 ⇒ cosθ =
    1
    5

    On subtracting,
    2tanθ= 2 + √5 - √5 + 2 = 4
    ⇒ tan θ = 2
    tanθ
    = sin θ
    2
    sec θ5

    ∴ sin θ + cos θ
    2
    +
    1
    55

    =
    3
    5


  1. If sin θ + cosec θ = 2, then the value of sin9θ + cosec9θ is :









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    sin θ + cosec θ = 2

    ⇒ sin θ =
    1
    = 2
    sin θ

    ⇒ sin² θ - 2 sinθ + 1 = 0
    ⇒ (sin θ - 1)² = 0
    ⇒ sin θ = 1
    ∴ cosec θ = 1
    ∴ sin9 θ + cosec9 θ = 1 + 1 = 2

    Correct Option: B

    sin θ + cosec θ = 2

    ⇒ sin θ =
    1
    = 2
    sin θ

    ⇒ sin² θ - 2 sinθ + 1 = 0
    ⇒ (sin θ - 1)² = 0
    ⇒ sin θ = 1
    ∴ cosec θ = 1
    ∴ sin9 θ + cosec9 θ = 1 + 1 = 2



  1. tan θ
    +
    cot θ
    is equal to
    1 - cot θ1 - tan θ









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    Expression

    =
    tan θ
    +
    cot θ
    1 - cot θ1 - tan θ

    =
    tan θ
    +
    (1 / tan θ)
    {1 - (1 / tan θ) }1 - tan θ

    =
    tan² θ
    +
    1
    tan θ - 1tan θ(1 - tan θ)

    =
    tan² θ
    +
    1
    tan θ - 1tan θ(tan θ - 1)

    =
    tan3 θ - 1
    tan θ (tan θ - 1)

    =
    (tan θ - 1)(tan ²θ + tan θ + 1)
    tan θ (tan θ - 1)

    =
    tan ²θ + tan θ + 1
    tan θ

    = tan θ + cot θ + 1

    Correct Option: D

    Expression

    =
    tan θ
    +
    cot θ
    1 - cot θ1 - tan θ

    =
    tan θ
    +
    (1 / tan θ)
    {1 - (1 / tan θ) }1 - tan θ

    =
    tan² θ
    +
    1
    tan θ - 1tan θ(1 - tan θ)

    =
    tan² θ
    +
    1
    tan θ - 1tan θ(tan θ - 1)

    =
    tan3 θ - 1
    tan θ (tan θ - 1)

    =
    (tan θ - 1)(tan ²θ + tan θ + 1)
    tan θ (tan θ - 1)

    =
    tan ²θ + tan θ + 1
    tan θ

    = tan θ + cot θ + 1


  1. If tan θ + cot θ = 2, then the value of tan100θ + cot100θ is









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    tan θ + cot θ = 2

    ⇒ tan θ +
    1
    = 2
    tan θ

    ⇒ tan²θ + 1 = 2 tanθ
    ⇒ tan²θ – 2tan θ + 1 = 0
    ⇒ (tanθ – 1)² = 0
    ⇒ tan θ = 1
    ∴ cot θ =
    1
    = 1
    tan θ

    ∴ tan100 θ + cot100 θ = 1 + 1 = 2

    Correct Option: A

    tan θ + cot θ = 2

    ⇒ tan θ +
    1
    = 2
    tan θ

    ⇒ tan²θ + 1 = 2 tanθ
    ⇒ tan²θ – 2tan θ + 1 = 0
    ⇒ (tanθ – 1)² = 0
    ⇒ tan θ = 1
    ∴ cot θ =
    1
    = 1
    tan θ

    ∴ tan100 θ + cot100 θ = 1 + 1 = 2



  1. If tan θ = 2, then the value of
    8sinθ + 5cosθ
    is :
    sin3θ + 2cos3θ + 3 cosθ









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    Expression

    =
    8 sin θ + 5 cos θ
    sin3θ + 2cos3θ + 3cos θ

    Dividing numerator and denominator by cos θ,
    =
    8 tan θ + 5
    tan θ . sin²θ + 2cos²θ + 3

    =
    8 tan θ + 5
    2sin² θ + 2cos²θ + 3

    =
    8 tan θ + 5
    2(sin² θ + 2cos²θ) + 3

    =
    8 × 2 + 5
    =
    21
    55

    Correct Option: A

    Expression

    =
    8 sin θ + 5 cos θ
    sin3θ + 2cos3θ + 3cos θ

    Dividing numerator and denominator by cos θ,
    =
    8 tan θ + 5
    tan θ . sin²θ + 2cos²θ + 3

    =
    8 tan θ + 5
    2sin² θ + 2cos²θ + 3

    =
    8 tan θ + 5
    2(sin² θ + 2cos²θ) + 3

    =
    8 × 2 + 5
    =
    21
    55