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Math Help - Verifying trig identities

  1. #1
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    Verifying trig identities

    Hello,

    I'm having difficulties verifying this problem, can some please help me out?

    Problem: (tan-cot) / (tan^2) - (cot^2) = sin cos
    if I'm right the problem should look like this...

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

    what are the algebraic steps to verify this problem?


    Thanks!
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by l flipboi l View Post
    Hello,

    I'm having difficulties verifying this problem, can some please help me out?

    Problem: (tan-cot) / (tan^2) - (cot^2) = sin cos
    if I'm right the problem should look like this...

    ((sinA/cosA)-(cosA/sinA)) / ((sin^2A/cos^2A)-(cos^2A/sin^2A))

    what are the algebraic steps to verify this problem?


    Thanks!
    Yep, so far so good. I would then cross multiply to give only one denominator:

    \frac{sinA}{cosA} - \frac{cosA}{sinA} = \frac{sin^2A-cos^2A}{sinAcosA}

    \frac{sin^2A}{cos^2A} - \frac{cos^2A}{sin^2A} = \frac{sin^4A-cos^4A}{sin^2Acos^2A}

    Therefore we get:

    \frac{sin^2A-cos^2A}{sinAcosA} \times \frac{sin^2Acos^2A}{sin^4A-cos^4A}.

    To remove the quartic recall the difference of two squares and that sin^2A+cos^2A = 1
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    Yep, so far so good. I would then cross multiply to give only one denominator:

    \frac{sinA}{cosA} - \frac{cosA}{sinA} = \frac{sin^2A-cos^2A}{sinAcosA}

    \frac{sin^2A}{cos^2A} - \frac{cos^2A}{sin^2A} = \frac{sin^4A-cos^4A}{sin^2Acos^2A}

    Therefore we get:

    \frac{sin^2A-cos^2A}{sinAcosA} \times \frac{sin^2Acos^2A}{sin^4A-cos^4A}.

    To remove the quartic recall the difference of two squares and that sin^2A+cos^2A = 1
    Thanks for the help! I was having difficulties with dividing fractions
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