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How do I prove this trigonometric identity: (tanΘ+secΘ-1)/(tanΘ-secΘ+... Walk me through it?

Update:

The problem didn't show clearly, but it's this: (tanΘ+secΘ-1)/(tanΘ-secΘ+1)=(tanΘ+secΘ)

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  • Puggy
    Lv 7
    1 decade ago
    Favorite Answer

    [tan(t) + sec(t) - 1] / [tan(t) - sec(t) + 1] = tan(t) + sec(t)

    To solve this, we choose the more complex side. In this case, it's the left hand side (LHS).

    LHS = [tan(t) + sec(t) - 1] / [tan(t) - sec(t) + 1]

    Change everything to sines and cosines.

    LHS = [sin(t)/cos(t) + 1/cos(t) - 1] / [sin(t)/cos(t) - 1/cos(t) + 1]

    Now, multiply everything by cos(t), to get rid of all fractions within fractions.

    LHS = [sin(t) + 1 - cos(t)] / [sin(t) - 1 + cos(t)]

    Let's rearrange the terms a bit.

    LHS = [{sin(t) - cos(t)} + 1] / [{sin(t) + cos(t)} - 1]

    Let's multiply the top and bottom by the conjugate of the bottom. That is, [(sin(t) + cos(t)) + 1]. Note that this will lead to a difference of squares in the denominator.

    LHS = [{sin(t) - cos(t)} + 1] [(sin(t) + cos(t)) + 1] /

    [(sin(t) + cos(t))^2 - 1]

    Simplify the denominator by expanding out the squared binomial, to get

    LHS = [{sin(t) - cos(t)} + 1] [(sin(t) + cos(t)) + 1] /

    [sin^2(t) + 2sin(t)cos(t) + cos^2(t) - 1)

    Note that sin^2(t) + cos^2(t) = 1, so

    LHS = [{sin(t) - cos(t)} + 1] [(sin(t) + cos(t)) + 1] /

    [2sin(t)cos(t) + 1 - 1]

    Simplfying some more,

    LHS = [{sin(t) - cos(t)} + 1] [(sin(t) + cos(t)) + 1] / [2sin(t)cos(t)]

    Now, let's expand the numerator.

    LHS = { [sin(t) - cos(t)][sin(t) + cos(t)] + [sin(t) - cos(t)] + [sin(t) + cos(t)] + 1 } / [2sin(t)cos(t)]

    Now, we simplify the numerator.

    LHS = { sin^2(t) - cos^2(t) + 2sin(t) + 1 } / [2sin(t)cos(t)]

    Notice we have a 1, and a -cos^2(t) in the numerator. This translates to 1 - cos^2(t), and can be changed to sin^2(t).

    LHS = { sin^2(t) + sin^2(t) + 2sin(t) } / [2sin(t)cos(t)]

    Simplify the top,

    LHS = [2sin^2(t) + 2sin(t)] / [2sin(t)cos(t)]

    Factor the top

    LHS = 2sin(t) [sin(t) + 1] / [2sin(t)cos(t)]

    The 2sin(t) on the top and bottom cancel.

    LHS = (sin(t) + 1) / cos(t)

    Now, split this up into two fractions.

    LHS = sin(t)/cos(t) + 1/cos(t)

    By definition,

    LHS = tan(t) + sec(t) = RHS

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