Trigonometry


  1. If tan θ =
    3
    , then the value of
    4sin²θ - 2cos²θ
    is equal to
    44sin²θ + 3cos²θ









  1. View Hint View Answer Discuss in Forum

    tan θ =
    3
    ⇒ tan² θ =
    9
    416

    Expression
    =
    4sin² θ - 2 cos² θ
    4sin² θ - 3 cos² θ


    =
    4tan² θ - 2
    4tan² θ + 3


    Correct Option: A

    tan θ =
    3
    ⇒ tan² θ =
    9
    416

    Expression
    =
    4sin² θ - 2 cos² θ
    4sin² θ - 3 cos² θ


    =
    4tan² θ - 2
    4tan² θ + 3



  1. If sin (θ + 18°) = cos 60° (0 < θ < 90°), then the value of cos 5θ is









  1. View Hint View Answer Discuss in Forum

    sin (θ + 18°) = cos 60°
    = cos (90° – 30°) = sin 30°
    ⇒ θ + 18° = 30°
    ⇒ θ = 30° – 18° = 12°

    ∴ cos5θ = cos 60° =
    1
    2

    Correct Option: A

    sin (θ + 18°) = cos 60°
    = cos (90° – 30°) = sin 30°
    ⇒ θ + 18° = 30°
    ⇒ θ = 30° – 18° = 12°

    ∴ cos5θ = cos 60° =
    1
    2



  1. The value of
    sin 25° cos 65° + cos 25° sin 65°
    tan²70° - cosec²20°









  1. View Hint View Answer Discuss in Forum

    sin25°cos65° + cos25°sin65°
    tan² 70° - cosec² 20°

    sin25°cos(90° - 25°) + cos25°
    sin(90° - sin25°)
    tan² 70° - cosec² (90° - 70°)


    sin25°sin25° + cos25°.cos25°
    tan² 70° - sec² 70°

    sin²25° + cos²25°
    tan² 70° - sec² 70°

    =
    1
    = - 1
    - 1

    Correct Option: A

    sin25°cos65° + cos25°sin65°
    tan² 70° - cosec² 20°

    sin25°cos(90° - 25°) + cos25°
    sin(90° - sin25°)
    tan² 70° - cosec² (90° - 70°)


    sin25°sin25° + cos25°.cos25°
    tan² 70° - sec² 70°

    sin²25° + cos²25°
    tan² 70° - sec² 70°

    =
    1
    = - 1
    - 1


  1. If (r cos θ – √3 )² + (r sin θ –1)² = 0 then the value of
    r tan θ + sec θ
    is equal to
    r sec θ + tan θ









  1. View Hint View Answer Discuss in Forum

    (r cos θ – √3 )⇒ + (r sin θ –1)⇒ = 0
    ⇒ r cos θ – √3 = 0 and r sin θ – 1 = 0
    ⇒ r cos θ = √3 and r sin θ = 1
    ∴ r² cos²θ + r² sin²θ = 3 + 1
    ⇒ r² (sin²θ + cos²θ) = 4
    ⇒ r² = 4
    ⇒ r = 2

    ∴ tan θ =
    r sin θ
    =
    1
    r cos θ3

    and r cosθ = √3 ⇒ cosθ =
    3
    r

    ⇒ sec θ =
    r
    3


    =
    2r
    =
    2 × 2
    =
    4

    r² + 14 + 15

    Correct Option: A

    (r cos θ – √3 )⇒ + (r sin θ –1)⇒ = 0
    ⇒ r cos θ – √3 = 0 and r sin θ – 1 = 0
    ⇒ r cos θ = √3 and r sin θ = 1
    ∴ r² cos²θ + r² sin²θ = 3 + 1
    ⇒ r² (sin²θ + cos²θ) = 4
    ⇒ r² = 4
    ⇒ r = 2

    ∴ tan θ =
    r sin θ
    =
    1
    r cos θ3

    and r cosθ = √3 ⇒ cosθ =
    3
    r

    ⇒ sec θ =
    r
    3


    =
    2r
    =
    2 × 2
    =
    4

    r² + 14 + 15



  1. If q is a positive acute angle and 4 cos²θ – 4 cos θ + 1 = 0, then the value of tan (θ – 15° ) is equal to









  1. View Hint View Answer Discuss in Forum

    4cos²θ – 4cosθ + 1 = 0
    ⇒ (2 cosθ – 1)² = 0
    ⇒ 2 cosθ – 1 = 0
    ⇒ 2 cosθ = 1

    ⇒ cosθ =
    1
    = cos 60°
    2

    ⇒ θ = 60°
    ∴ tan (θ –15°) = tan (60° –15°) =
    tan 45° = 1

    Correct Option: B

    4cos²θ – 4cosθ + 1 = 0
    ⇒ (2 cosθ – 1)² = 0
    ⇒ 2 cosθ – 1 = 0
    ⇒ 2 cosθ = 1

    ⇒ cosθ =
    1
    = cos 60°
    2

    ⇒ θ = 60°
    ∴ tan (θ –15°) = tan (60° –15°) =
    tan 45° = 1