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anonymous

  • one year ago

Given tan A + tan B = 3x and tan A tan B = 2x^2 How to find tan A - tan B ?

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  1. anonymous
    • one year ago
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    @ganeshie8

  2. anonymous
    • one year ago
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    from your 2nd equation, \[\tan(B)=\frac{2x^2}{\tan(A)}\] Put this in 1st equation \[\tan(A)+\frac{2x^2}{\tan(A)}=3x\] Multiply by tan(A) \[\tan^2(A)+2x^2=3x.\tan(A)\]\[\tan^2(A)-3x.\tan(A)+2x^2=0\] Split \[-3x=-x-2x\] and simplify to get tan(A) in terms of x, use the value of tan(A) in terms of x to get tan(B) in terms of x then evaluate tan(A)-tan(B) in terms of x

  3. ParthKohli
    • one year ago
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    I swear to God I just saw this question on Math.SE.

  4. anonymous
    • one year ago
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    @Nishant_Garg uhh.. how can you split that way when you have 2 unknowns.. ?

  5. anonymous
    • one year ago
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    @ParthKohli Hi Professor, what is Math.S.E ? can you kindly help me out on this ? ><

  6. ParthKohli
    • one year ago
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    \[(\tan A - \tan B) ^2 = (\tan A + \tan B)^2 - 4\tan A \tan B \]

  7. ParthKohli
    • one year ago
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    Someone asked exactly the same question on another side, and the difference is only of a minute or two.

  8. anonymous
    • one year ago
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    The answer will still be in terms of x only?or am I missing something?

  9. anonymous
    • one year ago
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    @ParthKohli now thats a equation we dont see often :| @Nishant_Garg yup! answer in terms of x

  10. ParthKohli
    • one year ago
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    \[(\tan A - \tan B)^2 = (3x)^2 - 4(2x)^2 = x^2\]\[\tan A - \tan B =x \]

  11. ParthKohli
    • one year ago
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    You do understand what that equation means though, right?

  12. anonymous
    • one year ago
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    @ParthKohli oh yes i got it! thank you so much!! :D

  13. ParthKohli
    • one year ago
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    Note that the answer can also be \(-x\).

  14. anonymous
    • one year ago
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    Yep I got \[\pm x\] from my method

  15. ParthKohli
    • one year ago
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    It was pretty visible too. \[x + 2x = 3x\]and\[x\cdot 2x = 2x^2\]

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