Trigonometry


  1. If sinθ + cosecθ = 2, the value of sin100θ + cosec100θ is :









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    sinθ + cosecθ = 2

    ⇒ sinθ +
    1
    = 2
    sinθ

    ⇒ sin2θ + 1 = 2sinθ
    ⇒ sin2θ - 2sinθ + 1 = 0
    ⇒ (sinθ - 1)2 = 0
    ⇒ sinθ = 1
    ∴ cosecθ =
    1
    = 1
    sinθ

    ∴ sin100θ + cosec100θ = 1 + 1 = 2

    Correct Option: B

    sinθ + cosecθ = 2

    ⇒ sinθ +
    1
    = 2
    sinθ

    ⇒ sin2θ + 1 = 2sinθ
    ⇒ sin2θ - 2sinθ + 1 = 0
    ⇒ (sinθ - 1)2 = 0
    ⇒ sinθ = 1
    ∴ cosecθ =
    1
    = 1
    sinθ

    ∴ sin100θ + cosec100θ = 1 + 1 = 2


  1. The value of
    sin 65°
    is
    cos 25°










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    sin 65°
    =
    sin( 90° - 25° )
    cos 25°cos 25°

    sin 65°
    =
    cos25°
    = 1
    cos 25°cos 25°

    Correct Option: B

    sin 65°
    =
    sin( 90° - 25° )
    cos 25°cos 25°

    sin 65°
    =
    cos25°
    = 1
    cos 25°cos 25°



  1. If
    sinx
    +
    sinx
    = 4 ; and 0° < x < 90°, then find the value of x
    1 + cosx1 - cosx










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    sinx
    +
    sinx
    = 4
    1 + cosx1 - cosx

    sinx(1 - cosx) + sinx(1 + cosx)
    = 4
    (1 + cosx)(1 - cosx)

    sinx - cosx.sinx + sinx + cosx.sinx
    = 4
    (1 - cos2x)

    2sinx
    = 4
    sin2x

    ⇒ 2sinx = 1
    ⇒ sinx =
    1
    = sin30° ⇒ x = 30°
    2

    Correct Option: D

    sinx
    +
    sinx
    = 4
    1 + cosx1 - cosx

    sinx(1 - cosx) + sinx(1 + cosx)
    = 4
    (1 + cosx)(1 - cosx)

    sinx - cosx.sinx + sinx + cosx.sinx
    = 4
    (1 - cos2x)

    2sinx
    = 4
    sin2x

    ⇒ 2sinx = 1
    ⇒ sinx =
    1
    = sin30° ⇒ x = 30°
    2


  1. Find the value of tan θ(1 + sec2θ)(1 + sec 4θ) (1 + sec8θ).









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    tan θ(1 + sec2θ)(1 + sec 4θ) (1 + sec8θ) = tanθ 1 +
    1
    1 +
    1
    1 +
    1
    cos2θcos4θcos8θ

    tan θ(1 + sec2θ)(1 + sec 4θ) (1 + sec8θ) = tanθ
    cos2θ + 1
    cos4θ + 1
    cos8θ + 1
    cos2θcos4θcos8θ

    tan θ(1 + sec2θ)(1 + sec 4θ) (1 + sec8θ) = tanθ
    2cos2θ
    2cos2
    2cos2
    cos2θcos4θcos8θ

    [ ∴ 1 + cos2θ = 2cos2θ ]
    Required answer = 8 .
    tanθ.cos2θ .cos2θ.cos4θ
    cos8θ

    Required answer = 4 .
    2sinθcosθ.cos2θ.cos4θ
    cos8θ

    Required answer = 4 .
    sin2θ.cos2θ.cos4θ
    cos8θ

    Required answer = 2 .
    2sin2θ.cos2θ.cos4θ
    cos8θ

    Required answer =
    2sin4θ.cos4θ
    cos8θ

    Required answer =
    sin8θ
    = tan8θ
    cos8θ

    Correct Option: B

    tan θ(1 + sec2θ)(1 + sec 4θ) (1 + sec8θ) = tanθ 1 +
    1
    1 +
    1
    1 +
    1
    cos2θcos4θcos8θ

    tan θ(1 + sec2θ)(1 + sec 4θ) (1 + sec8θ) = tanθ
    cos2θ + 1
    cos4θ + 1
    cos8θ + 1
    cos2θcos4θcos8θ

    tan θ(1 + sec2θ)(1 + sec 4θ) (1 + sec8θ) = tanθ
    2cos2θ
    2cos2
    2cos2
    cos2θcos4θcos8θ

    [ ∴ 1 + cos2θ = 2cos2θ ]
    Required answer = 8 .
    tanθ.cos2θ .cos2θ.cos4θ
    cos8θ

    Required answer = 4 .
    2sinθcosθ.cos2θ.cos4θ
    cos8θ

    Required answer = 4 .
    sin2θ.cos2θ.cos4θ
    cos8θ

    Required answer = 2 .
    2sin2θ.cos2θ.cos4θ
    cos8θ

    Required answer =
    2sin4θ.cos4θ
    cos8θ

    Required answer =
    sin8θ
    = tan8θ
    cos8θ



  1. If tanθ1 = 1, sinθ2 =
    1
    , then the value of sin (θ1 + θ2) equal to
    2










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    tanθ1 = 1 = tan 45°
    ⇒ θ1 = 45°

    Again , sinθ2 =
    1
    = sin45°
    2

    ⇒ θ2 = 45°
    ∴ sin(θ1 + θ2) = sin90° = 1

    Correct Option: C

    tanθ1 = 1 = tan 45°
    ⇒ θ1 = 45°

    Again , sinθ2 =
    1
    = sin45°
    2

    ⇒ θ2 = 45°
    ∴ sin(θ1 + θ2) = sin90° = 1