Trigonometry


  1. If tan4θ + tan2θ = 1 then the value of cos4θ + cos2θ is









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    tan4θ + tan2θ = 1
    ⇒  tan2θ (tan2θ + 1) = 1
    ⇒  tan2θ . sec2θ = 1

    ⇒  tan2θ =
    1
    = cos2θ.
    sec2θ

    ∴  cos4θ + cos2θ = tan4θ + tan2θ = 1

    Correct Option: C

    tan4θ + tan2θ = 1
    ⇒  tan2θ (tan2θ + 1) = 1
    ⇒  tan2θ . sec2θ = 1

    ⇒  tan2θ =
    1
    = cos2θ.
    sec2θ

    ∴  cos4θ + cos2θ = tan4θ + tan2θ = 1


  1. (1 + sec 20° + cot 70°)(1 – cosec 20° + tan70°) is equal to









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    (1 + sec 20° + cot 70°) (1 – cosec 20° + tan 70°)
    = (1 + sec 20° + tan 20°) (1 – cosec 20° + cot 20°)
    [∴  tan (90° – θ) = cotθ; cot (90° – θ) = tanθ]

    = 1 +
    1
    +
    sin20°
    1 −
    1
    +
    cos20°
    cos20°cos20°sin20°sin20°

    =
    cos20° + 1 + sin20°
    sin20° – 1 + cos20°
    cos20°sin20°

    =
    (sin20° + cos20° + 1)( sin20° + cos20° + 1)
    sin20°.cos20°

    =
    (sin20° + cos20°)2 − 1
    sin20°.cos20°

    =
    sin220° + cos220° + 2sin20°.cos20°− 1
    sin20°.cos20°

    =
    1 + 2sin20°.cos20°− 1
    = 2
    sin20°.cos20°

    Correct Option: C

    (1 + sec 20° + cot 70°) (1 – cosec 20° + tan 70°)
    = (1 + sec 20° + tan 20°) (1 – cosec 20° + cot 20°)
    [∴  tan (90° – θ) = cotθ; cot (90° – θ) = tanθ]

    = 1 +
    1
    +
    sin20°
    1 −
    1
    +
    cos20°
    cos20°cos20°sin20°sin20°

    =
    cos20° + 1 + sin20°
    sin20° – 1 + cos20°
    cos20°sin20°

    =
    (sin20° + cos20° + 1)( sin20° + cos20° + 1)
    sin20°.cos20°

    =
    (sin20° + cos20°)2 − 1
    sin20°.cos20°

    =
    sin220° + cos220° + 2sin20°.cos20°− 1
    sin20°.cos20°

    =
    1 + 2sin20°.cos20°− 1
    = 2
    sin20°.cos20°



  1. If a2 sec2x – b2 tan2x = c2 then the
    value of (sec2x + tan2x) is equal to (assume b2 ≠ a2)









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    a2 sec2x – b2 tan2x = c2
    ⇒  a2 (1 + tan2x) – b2 tan2x = c2
    ⇒  a2 + a2tan2x – b2 tan2x = c2
    ⇒  a2tan2x – b2 tan2x = c2 – a2
    ⇒  tan2x(a2 – b2) = c2 – a2

    ⇒  tan2x =
    c2 – a2
    a2 – b2

    ∴  sec2x + tan2x
    = 1 + tan2x + tan2x
    = 1 + 2 tan2x
    = 1 +
    2(c2 – a2)
    a2 – b2

    =
    a2 – b2 + 2c2 – 2a2
    a2 – b2

    =
    – b2 + 2c2 – a2
    a2 – b2

    =
    b2 + a2 – 2c2
    b2 – a2

    Correct Option: B

    a2 sec2x – b2 tan2x = c2
    ⇒  a2 (1 + tan2x) – b2 tan2x = c2
    ⇒  a2 + a2tan2x – b2 tan2x = c2
    ⇒  a2tan2x – b2 tan2x = c2 – a2
    ⇒  tan2x(a2 – b2) = c2 – a2

    ⇒  tan2x =
    c2 – a2
    a2 – b2

    ∴  sec2x + tan2x
    = 1 + tan2x + tan2x
    = 1 + 2 tan2x
    = 1 +
    2(c2 – a2)
    a2 – b2

    =
    a2 – b2 + 2c2 – 2a2
    a2 – b2

    =
    – b2 + 2c2 – a2
    a2 – b2

    =
    b2 + a2 – 2c2
    b2 – a2


  1. x, y be two acute angles, x + y < 90° and sin(2x – 20°) = cos (2y + 20°), the value of tan (x + y) is









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    sin (2x – 20°) = cos (2y + 20°)
    ⇒  sin (2x – 20°)
    = sin {90° – (2y + 20°)}
    ⇒  2x – 20° = 90° – 2y – 20°
    ⇒  2x + 2y = 90°
    ⇒  2 (x + y) = 90° ⇒ x + y = 45°
    ∴  tan (x + y) = tan 45° = 1

    Correct Option: C

    sin (2x – 20°) = cos (2y + 20°)
    ⇒  sin (2x – 20°)
    = sin {90° – (2y + 20°)}
    ⇒  2x – 20° = 90° – 2y – 20°
    ⇒  2x + 2y = 90°
    ⇒  2 (x + y) = 90° ⇒ x + y = 45°
    ∴  tan (x + y) = tan 45° = 1



  1. If r sinθ = √3 r cosθ = 1, rhen values of r and θ are : (0° ≤ θ ≤ 90°)









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    r sinθ = √3
    r cosθ = 1
    On squaring and adding,
    r2sin2θ + r2cos2θ = 3 + 1
    ⇒  r2 (sin2θ + cos2θ) = 4

    Again, 
    r sinθ
    = √3
    r cosθ

    ⇒  tanθ = √3 = tan 60°
    ⇒  θ = 60°

    Correct Option: D

    r sinθ = √3
    r cosθ = 1
    On squaring and adding,
    r2sin2θ + r2cos2θ = 3 + 1
    ⇒  r2 (sin2θ + cos2θ) = 4

    Again, 
    r sinθ
    = √3
    r cosθ

    ⇒  tanθ = √3 = tan 60°
    ⇒  θ = 60°