Trigonometry....sine, cosine, tangent question?

Verify that the equation is an identity

( (tan^2 x) / (csc^2 x + sec^2 x) ) = sin^2 x - sin^2 x cos^2 x

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  • Anonymous
    8 years ago
    Favorite Answer

    Put the whole left side (LHS) into sines and cosines:

    tan^2 = sin^2 / cos^2

    csc^2 = sin^-2

    sec^2 = cos^-2

    rearrange the denominator:

    sin^-2 + cos^-2 =(cos^2 + sin^2)/(sin^2 * cos^2)

    LHS = tan^2 / ((cos^2 + sin^2)/(sin^2 * cos^2)) = (sin^2 * (sin^2 * cos^2)) / (cos^2 *(cos^2 + sin^2))

    cancel the cos^2 and rearrange:

    LHS = (sin^4) / (cos^2 + sin^2)

    sin^2 + cos^2 = 1, so LHS = sin^4

    Now, look at the right-hand side (RHS)

    RHS = sin^2 - sin^2*cos^2 = sin^2*(1 - cos^2)

    sin^2 + cos^2 = 1, so 1-cos^2 = sin^2, so the RHS is also equal to sin^4

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