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Aptitude miscellaneous
  1. Find numerical value of
    9
    		+ 4 cos² θ +
    5
    cosec² θ1 + tan² θ








  1. View Hint View Answer Discuss in Forum

    9
    		+ 4 cos² θ +
    5
    		= 90
    cosec² θ1 + tan² θ

    = 9 sin² θ + 4 cos² θ +
    5
    sec² θ

    = 9 sin² θ + 4 cos² θ + 5 cos² θ
    = 9 sin² θ + 9 cos² θ
    = 9 (sin² θ + cos² θ) = 9 × 1 = 9

    Correct Option: C

    9
    		+ 4 cos² θ +
    5
    		= 90
    cosec² θ1 + tan² θ

    = 9 sin² θ + 4 cos² θ +
    5
    sec² θ

    = 9 sin² θ + 4 cos² θ + 5 cos² θ
    = 9 sin² θ + 9 cos² θ
    = 9 (sin² θ + cos² θ) = 9 × 1 = 9

  1. If tanθ + secθ = 3, θ being acute, the value of 5 sinθ is :








  1. View Hint View Answer Discuss in Forum

    tanθ + secθ = 3 ......(i)
    ∵ sec² θ – tan² θ = 1
    ⇒ (secθ – tanθ) (secθ + tanθ) = 1

    ⇒ secθ – tanθ =
    1
    		 .... (ii)
    3

    On adding equations (i) and (ii),
    2 secθ = 3 +
    1
    		=
    10
    33

    On subtracting equation (ii) from (i),
    2tanθ = 3 -
    1
    3

    =
    9 - 1
    		=
    8
    33

    ∴ sin θ =
    tan θ
    sec θ

    =
    8
    		×
    3
    		=
    4

    3105

    ∴ 5 sinθ = 5 ×
    4
    		= 4
    5

    Correct Option: D

    tanθ + secθ = 3 ......(i)
    ∵ sec² θ – tan² θ = 1
    ⇒ (secθ – tanθ) (secθ + tanθ) = 1

    ⇒ secθ – tanθ =
    1
    		 .... (ii)
    3

    On adding equations (i) and (ii),
    2 secθ = 3 +
    1
    		=
    10
    33

    On subtracting equation (ii) from (i),
    2tanθ = 3 -
    1
    3

    =
    9 - 1
    		=
    8
    33

    ∴ sin θ =
    tan θ
    sec θ

    =
    8
    		×
    3
    		=
    4

    3105

    ∴ 5 sinθ = 5 ×
    4
    		= 4
    5

  1. If cosθ =
    p
    		then the value of tanθ is :
    √p² + q²








  1. View Hint View Answer Discuss in Forum

    cosθ =
    p
    √p² + q²

    ∴ sinθ = √1 - cos² θ

    ∴ tanθ =
    sinθ
    cosθ

    =
    q
    		×
    √p² + q²
    		=
    q
    √p² + q²pp

    Correct Option: B

    cosθ =
    p
    √p² + q²

    ∴ sinθ = √1 - cos² θ

    ∴ tanθ =
    sinθ
    cosθ

    =
    q
    		×
    √p² + q²
    		=
    q
    √p² + q²pp

  1. If θ be acute angle and tan (4θ – 50°) = cot(50° – θ), then the value of θ in degrees is :








  1. View Hint View Answer Discuss in Forum

    tan (4θ – 50°) = cot (50° – θ)
    ⇒ tan (4θ – 50°)
    = tan (90° – (50° – θ))
    ⇒ 4θ – 50° = 90° – (50° – θ)
    ⇒ 4θ – 50° = 90° – 50° + θ
    ⇒ 4θ – 50° = 40° + θ
    ⇒ 4θ – θ = 40° + 50°

    ⇒ 3θ = 90° ⇒ θ =
    90°
    		= 30°
    3

    Correct Option: D

    tan (4θ – 50°) = cot (50° – θ)
    ⇒ tan (4θ – 50°)
    = tan (90° – (50° – θ))
    ⇒ 4θ – 50° = 90° – (50° – θ)
    ⇒ 4θ – 50° = 90° – 50° + θ
    ⇒ 4θ – 50° = 40° + θ
    ⇒ 4θ – θ = 40° + 50°

    ⇒ 3θ = 90° ⇒ θ =
    90°
    		= 30°
    3

  1. If sin θ + sin²θ = 1 then cos²θ + cos4θ is equal to








  1. View Hint View Answer Discuss in Forum

    sinθ + sin²θ = 1
    ⇒ sinθ = 1 – sin²θ = cos2θ
    ∴ cos²θ + cos4θ
    = cos²θ + (cos²θ)²
    = cos²θ + sin²θ = 1

    Correct Option: B

    sinθ + sin²θ = 1
    ⇒ sinθ = 1 – sin²θ = cos2θ
    ∴ cos²θ + cos4θ
    = cos²θ + (cos²θ)²
    = cos²θ + sin²θ = 1