Trigonometry


  1. If θ is a positive acute angle and cosec θ = √3 , then the value of cot θ – cosec θ is











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    cosecθ = √3
    cotθ = √cosec² θ - 1
    =(√3² - 1) = √3 - 1 = √2
    ∴ cotθ - cosecθ = √2 - √3

    =
    3√2 - √2
    = (√2 - √3)
    3

    Correct Option: E

    cosecθ = √3
    cotθ = √cosec² θ - 1
    =(√3² - 1) = √3 - 1 = √2
    ∴ cotθ - cosecθ = √2 - √3

    =
    3√2 - √2
    = (√2 - √3)
    3


  1. If α and β are positive acute angles, sin (4α – β) = 1 and cos (2 α + β) = (1 / 2) , then the value of sin (α + 2β) is









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    sin (4α – β) = 1 = sin 90°
    ⇒ 4α – β = 90° ...(i)

    cos (2α + β) =
    1
    = cos 60°
    2

    ⇒ 2α + β = 60° ...(ii)
    On adding equations (i) and (ii),
    4α - β + 2α + β = 90° + 60°
    ⇒ 6α = 150° ⇒ α =
    150
    = 25°
    6

    From equation (ii),
    2 × 25 + β = 60°
    ⇒ β = 60° – 50° = 10°
    ∴ sin (α + 2β)
    = sin (25 + 2 × 10)
    = sin 45° =
    1
    = 25°
    2

    Correct Option: D

    sin (4α – β) = 1 = sin 90°
    ⇒ 4α – β = 90° ...(i)

    cos (2α + β) =
    1
    = cos 60°
    2

    ⇒ 2α + β = 60° ...(ii)
    On adding equations (i) and (ii),
    4α - β + 2α + β = 90° + 60°
    ⇒ 6α = 150° ⇒ α =
    150
    = 25°
    6

    From equation (ii),
    2 × 25 + β = 60°
    ⇒ β = 60° – 50° = 10°
    ∴ sin (α + 2β)
    = sin (25 + 2 × 10)
    = sin 45° =
    1
    = 25°
    2



  1. If 0° < A < 90°, then the value of tan²A + cot² A – sec² A cosec² A is









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    tan²A + cot²A – sec²A . cosec²A
    = tan²A + cot²A – (1 + tan²A) (1 + cot²A)
    = tan²A + cot²A – (1 + tan²A +
    cot²A + cot²A.tan²A)
    = tan²A + cot²A – 1 – tan²A – cot²A – cot²A . tan²A
    = – 1 – 1 = – 2
    [ tanA . cotA = 1]

    Correct Option: D

    tan²A + cot²A – sec²A . cosec²A
    = tan²A + cot²A – (1 + tan²A) (1 + cot²A)
    = tan²A + cot²A – (1 + tan²A +
    cot²A + cot²A.tan²A)
    = tan²A + cot²A – 1 – tan²A – cot²A – cot²A . tan²A
    = – 1 – 1 = – 2
    [ tanA . cotA = 1]


  1. If cos θ =
    3
    then the value of sinθ . secθ . tanθ is
    5









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    cos θ =
    3
    5

    ∴ sec θ =
    5
    3

    ∴ tan θ = √sec² θ - 1

    =
    4
    3

    ∴ sinθ . secθ . tanθ =
    sin θ
    . tan θ
    cos θ

    = tan²θ =4 ² =16
    39

    Correct Option: B

    cos θ =
    3
    5

    ∴ sec θ =
    5
    3

    ∴ tan θ = √sec² θ - 1

    =
    4
    3

    ∴ sinθ . secθ . tanθ =
    sin θ
    . tan θ
    cos θ

    = tan²θ =4 ² =16
    39












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    Expression

    =
    1 + sin θ
    +
    1 - sin θ
    cos θcos θ

    =
    1 + sin θ + 1 - sin θ
    +
    2
    cos θcos θ

    = 2 secθ

    Correct Option: D

    Expression

    =
    1 + sin θ
    +
    1 - sin θ
    cos θcos θ

    =
    1 + sin θ + 1 - sin θ
    +
    2
    cos θcos θ

    = 2 secθ