Trigonometry?

Prove that,

(Cosec A + Cot A)(1-Cos A) -(Sec A + Tan A)(1 - Sin A) = (Cosec A - Sec A)[2 - (1 - Cos A)(1- Sin A)]

Update:

Prove that,

(Cosec A + Cot A)(1-Sin A) -(Sec A + Tan A)(1 - Cos A) = (Cosec A - Sec A)[2 - (1 - Cos A)(1- Sin A)]

5 Answers

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  • 2 years ago
    Favorite Answer

     

    This is not an identity. Either you entered it incorrectly, or you meant for us to "solve" the equation, instead of "proving" it.

    LHS

    = (csc A + cot A) (1 − cos A) − (sec A + tan A) (1 − sin A)

    = (1/sin A + cos A/sin A) (1 − cos A) − (1/cos A + sin A/cos A) (1 − sin A)

    = (1 + cos A) (1 − cos A) / sin A − (1 + sin A) (1 − sin A) / cos A

    = (1 − cos²A) / sin A − (1 − sin²A) / cos A

    = sin²A / sin A − cos²A / cos A

    = sin A − cos A

    RHS

    = (csc A − sec A) [2 − (1 − cos A)(1 − sin A)]

    = (1/sin A − 1/cos A) [2 − (1 − cos A − sin A + sin A cos A)]

    = (cos A − sin A) (1 + cos A + sin A − sin A cos A) / (sin A cos A)

    LHS = RHS

    sin A − cos A = (cos A − sin A) (1 + cos A + sin A − sin A cos A) / (sin A cos A)

    (sin A − cos A) (sin A cos A) = (cos A − sin A) (1 + cos A + sin A − sin A cos A)

    (sin A − cos A) (sin A cos A) − (cos A − sin A) (1 + cos A + sin A − sin A cos A) = 0

    (sin A − cos A) (sin A cos A) + (sin A − cos A) (1 + cos A + sin A − sin A cos A) = 0

    (sin A − cos A) (sin A cos A + 1 + cos A + sin A − sin A cos A) = 0

    (sin A − cos A) (1 + cos A + sin A) = 0

    sin A − cos A = 0

    sin A = cos A

    sin A/cos A = 1

    tan A = 1

    A = π/4 + πk

    1 + cos A + sin A = 0

    cos A + sin A = −1

    cos A = −1 (and sin A = 0) or sin A = −1 (and cos A = 0)

    But restrictions on original equation does not allow sin A = 0 or cos A = 0

    A = π/4 + πk

  • 2 years ago

    Thanks to you all, the textbook made mistake in the question

  • 2 years ago

    (csc(A) + cot(A))(1 - cos(A)) - (sec(A) + tan(A))(1 - sin(A)) (LHS)

    = (csc(A) - cot(A) + cot(A) - cos(A)cot(A)) - (sec(A) - tan(A) + tan(A) - sin(A)tan(A))

    = csc(A) - cos(A)cot(A) - sec(A) + sin(A)tan(A)

    = (1/sin(A)) - (cos^2(A)/sin(A)) - (1/cos(A)) + (sin^2(A)/cos(A))

    = (cos(A) - cos^3(A) - sin(A) + sin^3(A)) / sin(A)cos(A)

    = (cos(A)(1 - cos^2(A) - sin(A)(1 - sin^2(A)))) / sin(A)cos(A)

    = (cos(A)sin^2(A) - sin(A)cos^2(A)) / sin(A)cos(A)

    = (sin(A)cos(A)(sin(A) - cos(A))) / sin(A)cos(A)

    = sin(A) - cos(A). Not RHS.

    Not an identity. (This only works for x = n * pi + (1/4)pi, assuming n is an integer.)

  • hfshaw
    Lv 7
    2 years ago

    You can't prove this because it's not true.

    (csc(A) + cot(A))(1 - cos(A)) - (sec(A) + tan(A))(1 - sin(A)) ?=? (csc(A) - sec(A))[2 - (1 - cos(A))(1 - sin(A))]

    (1/sin(A) + cos(A)/sin(A))*(1 - cos(A)) - (1/cos(A) + sin(A)/cos(A))*(1 - sin(A)) ?=? (1/sin(A) - 1/cos(A))*[2 - (1 - cos(A))(1 - sin(A)]

    (1 + cos(A))*(1 - cos(A))/sin(A) - (1 + sin(A))(1 - sin(A))/cos(A) ?=? (1/sin(A) - 1/cos(A))*[2 - (1 - cos(A))(1 - sin(A)]

    (1 - cos²(A))/sin(A) - (1 - sin²(A))/cos(A) ?=? (1/sin(A) - 1/cos(A))*[2 - (1 - cos(A))(1 - sin(A)]

    sin²(A)/sin(A) - cos²(A)/cos(A) ?=? (1/sin(A) - 1/cos(A))*[2 - (1 - cos(A))(1 - sin(A)]

    sin(A) - cos(A) ?=? (1/sin(A) - 1/cos(A))*[2 - (1 - cos(A))(1 - sin(A)]

    sin(A) - cos(A) ?=? (1/sin(A) - 1/cos(A))*[1 - sin(A)*cos(A) + cos(A) + sin(A)]

    sin(A) - cos(A) ?=? (cos(A) - sin(A))/(sin(A)*cos(A))*[1 - sin(A)*cos(A) + cos(A) + sin(A)]

    sin(A)*cos(A)*(sin(A) - cos(A)) ?=? (sin(A) - cos(A))*[sin(A)*cos(A) - cos(A) - sin(A) - 1]

    0 = cos(A) + sin(A) + 1

    cos(A) + sin(A) ≠ -1

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  • Anonymous
    2 years ago

    You don't bother to pick a Best Answer or say thanks (I checked your recent questions).

    So I won't be bothering to answer.

    • Am sorry about that, I will correct myself about that. Please, help me prove the above.

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