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Thread: Proof L.H.S=R.H.S

  1. #1
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    Proof L.H.S=R.H.S

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

    I want to show L.H.S=R.H.S. Please help me.
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  2. #2
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    Re: Proof L.H.S=R.H.S

    Quote Originally Posted by Dilan View Post
    (1/sec^2 A - cos^2 A + 1/cosec^2 A-sin^2 A) cos^2 A sin^2 A = (1-cos^2 A sin^2 A/2+cos^2 A sin^2 A)
    is this your equation?

    $\left(\dfrac{1}{\sec^2{A}} - \cos^2{A} + \dfrac{1}{\csc^2{A} - \sin^2{A}} \right)\cos^2{A}\sin^2{A} = \dfrac{1-\cos^2{A}\sin^2{A}}{2+\cos^2{A}\sin^2{A}}$

    If not, it helps if you make better use of brackets/parentheses to make numerators and denominators clear ...
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  3. #3
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    Re: Proof L.H.S=R.H.S

    There is serious difficulty with showing "L.H.S= R.H.S."- it isn't true!

    My first thought with anything like this is "change everything to sine and cosine". sec(A) is 1/cos(A) so that the first term, 1/sec^2(A), is just cos^2(A) and is immediately canceled by the second term, -cos^2(A). Similarly, cosec(A) is 1/sin(A) so 1/cosec^2(A)- sin^2(A)= 0 also. The quantity in the parentheses, multiplying cos^2(A)sin^2(A), is 0 so the left hand side is 0 for all A.

    But taking A= pi/2 or 90 degrees, cos(A)= 0 so both cos^2(A)sin^2(A/2) and cos^2(A)sin^2(A) are 0. The right hand side is 1 for A= pi/2 so the left hand side is NOT equal to the right hand side.
    Thanks from DenisB
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    Re: Proof L.H.S=R.H.S

    In fact, it is meant to be:

    $\left(\dfrac{1}{\sec^2{A} \ - \ \cos^2{A}} + \dfrac{1}{\csc^2{A} \ - \ \sin^2{A}} \right)\cos^2{A}\sin^2{A} \ = \ \dfrac{1-\cos^2{A}\sin^2{A}}{2+\cos^2{A}\sin^2{A}}$


    That revision is a correct identity. I worked it out.


    Dilan, your grouping symbols need to be present, for example:


    [1/(sec^2 A - cos^2 A) + 1/(cosec^2 A - sin^2 A)]cos^2 A*sin^2 A = (1 - cos^2 A*sin^2 A)/(2 + cos^2 A*sin^2 A)
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