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Math Help - Need help proving trig identity.

  1. #1
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    Exclamation Need help proving trig identity.

    I don't know how to star solving this problem, I'm new in calculus..

    f(x+y) - f(x) = sec^2 x tan y / 1 +tan x tan y
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  2. #2
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    What are u trying to find?
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  3. #3
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    this one sir..

    given: f(x)=tan x ,show that

    f(x+y) - f(x) = sec^2 x tan y / 1 +tan x tan y
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  4. #4
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    Whenever I have a trig identity to prove, I usually get everything in terms of sines and cosines. Use the addition of angles formulas. Then simplify.
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  5. #5
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    Quote Originally Posted by jasonlewiz View Post
    this one sir..

    given: f(x)=tan x ,show that

    f(x+y) - f(x) = sec^2 x tan y / 1 +tan x tan y
    the result should be \frac{\sec^2{x} \tan{y}}{1 - \tan{x}\tan{y}}
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  6. #6
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    As Ackbeet said, this is a matter of a trig identity and you can reduce to sine and cosine.

    tan(x+y)= \frac{sin(x+y)}{cos(x+y)}= \frac{sin(x)cos(y)+ cos(x)sin(y)}{cos(x)cos(y)- sin(x)sin(y)}

    Now, divide both numerator and denominator by cos(x)cos(y)
    tan(x+y)= \frac{\frac{sin(x)}{cos(x)}+ \frac{sin(y)}{cos(y)}}{1- \frac{sin(x)}{cos(x)}\frac{sin(y)}{cos(y)}}
    = \frac{tan(x)+ tan(y)}{1- tan(x)tan(y)}
    So that tan(x+y)- tan(x)= \frac{tan(x)+ tan(y)}{1- tan(x)tan(y)}- tan(x)
    = \frac{tan(x)+ tan(y)}{1- tan(x)tan(y)}- \frac{tan(x)- tan^2(x)tan(y)}{1- tan(x)tan(y)}

    Note that, as skeeter said, you have the sign in the denominator wrong.
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