Trigonometry


  1. If sec²θ + tan²θ = 7, then the value of θ when 0° ≤ θ ≤ 90°, is









  1. View Hint View Answer Discuss in Forum

    sec²θ + tan²θ = 7
    ⇒ 1 + tan²θ + tan²θ = 7
    ⇒ 2 tan²θ = 7 – 1 = 6
    ⇒ tan²θ = 3 ⇒ tanθ = 3
    ⇒ θ = 60°

    Correct Option: A

    sec²θ + tan²θ = 7
    ⇒ 1 + tan²θ + tan²θ = 7
    ⇒ 2 tan²θ = 7 – 1 = 6
    ⇒ tan²θ = 3 ⇒ tanθ = 3
    ⇒ θ = 60°


  1. If
    sin θ + cos θ
    = 3, then the value of sin4θ – cos4θ is
    sin θ - cos θ









  1. View Hint View Answer Discuss in Forum

    sin θ + cos θ
    = 3
    sin θ + cos θ

    ⇒ sinθ + cosθ = 3sinθ – 3 cosθ
    ⇒ 4cosθ = 2 sinθ ⇒ tanθ = 2
    ∴ sin4θ – cos4θ
    = (sin²θ + cos²θ) (sin²θ – cos²θ)
    = sin²θ– cos²θ
    = cos²θ (tan² θ – 1)
    =
    tan²θ - 1
    ²θ

    =
    tan²θ - 1
    =
    4 - 1
    =
    3

    1 + tan²θ1 + 45

    Correct Option: C

    sin θ + cos θ
    = 3
    sin θ + cos θ

    ⇒ sinθ + cosθ = 3sinθ – 3 cosθ
    ⇒ 4cosθ = 2 sinθ ⇒ tanθ = 2
    ∴ sin4θ – cos4θ
    = (sin²θ + cos²θ) (sin²θ – cos²θ)
    = sin²θ– cos²θ
    = cos²θ (tan² θ – 1)
    =
    tan²θ - 1
    ²θ

    =
    tan²θ - 1
    =
    4 - 1
    =
    3

    1 + tan²θ1 + 45



  1. If 2cosθ – sinθ = (1 / √2) , (0° < q < 90°) the value of 2 sinθ + cosθ is









  1. View Hint View Answer Discuss in Forum

    2 cosθ – sinθ =
    1
    2

    2sinθ + cosθ = x (Let)
    On squaring and adding,
    4cos²θ + sin²θ – 4 sinθ . cosθ + 4 sin²θ + cos²θ+ 4 sinθ.cosθ
    =
    1
    2

    1
    + x² = 5
    2

    ⇒x² = 5 -
    1
    =
    9
    ⇒ x =
    3

    222

    Correct Option: C

    2 cosθ – sinθ =
    1
    2

    2sinθ + cosθ = x (Let)
    On squaring and adding,
    4cos²θ + sin²θ – 4 sinθ . cosθ + 4 sin²θ + cos²θ+ 4 sinθ.cosθ
    =
    1
    2

    1
    + x² = 5
    2

    ⇒x² = 5 -
    1
    =
    9
    ⇒ x =
    3

    222


  1. If sin θ – cos θ = (7 / 13) and 0 < θ < 90°, then the value of sin θ + cos θ is









  1. View Hint View Answer Discuss in Forum

    sin θ – cos θ =
    7
    ......(i)
    13

    sin θ + cos θ = x .....(ii)
    On squaring both equations and adding,
    2(sin²θ + cos²θ) =
    49
    + x²
    169

    ⇒ x² = 2 -
    49
    =
    338 - 49
    169169

    =
    289
    ⇒ x =
    17
    16913

    Correct Option: A

    sin θ – cos θ =
    7
    ......(i)
    13

    sin θ + cos θ = x .....(ii)
    On squaring both equations and adding,
    2(sin²θ + cos²θ) =
    49
    + x²
    169

    ⇒ x² = 2 -
    49
    =
    338 - 49
    169169

    =
    289
    ⇒ x =
    17
    16913



  1. If θ be acute and tan θ + cot θ = 2, then the value of tan5 θ + cot10 θ is









  1. View Hint View Answer Discuss in Forum

    tan θ + cot θ = 2

    ⇒tan θ +
    1
    = 2
    tan θ

    ⇒ tan² θ + 1 = 2tan θ
    ⇒ tan² θ – 2tan θ + 1 = 0
    ⇒ (tan θ – 1)² = 0
    ⇒ tan θ = 1 ⇒ cot θ = 1
    ∴ tan5 θ + cot10 θ = 1 + 1 = 2

    Correct Option: B

    tan θ + cot θ = 2

    ⇒tan θ +
    1
    = 2
    tan θ

    ⇒ tan² θ + 1 = 2tan θ
    ⇒ tan² θ – 2tan θ + 1 = 0
    ⇒ (tan θ – 1)² = 0
    ⇒ tan θ = 1 ⇒ cot θ = 1
    ∴ tan5 θ + cot10 θ = 1 + 1 = 2