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AravindG

  • 2 years ago

Is this identity valid ?

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  1. AravindG
    • 2 years ago
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    \[\large \tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\frac{x+y+z-xyz}{1-xy-yz-zx}\]

  2. AravindG
    • 2 years ago
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    @UnkleRhaukus , @.Sam. , @Callisto

  3. mukushla
    • 2 years ago
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    let x=y=z=1

  4. AravindG
    • 2 years ago
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    so?

  5. mukushla
    • 2 years ago
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    lol.....im wrong...nothing.......lets think again

  6. AravindG
    • 2 years ago
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    @amistre64 , @experimentX

  7. experimentX
    • 2 years ago
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    no that's the right trick ... test for few arbitrary values.

  8. experimentX
    • 2 years ago
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    that's how i validate things ... before doing it if it looks ugly.

  9. AravindG
    • 2 years ago
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    i jst got to this eqn by myseslf ... so i dont knw ifthis can be generalised for all x,y,q

  10. AravindG
    • 2 years ago
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    i didnt see such an identituy in any textbook, i only saw tan-1 x+tan-1 y

  11. AravindG
    • 2 years ago
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    can anyone tell me if this is valid for all x,y ,z?

  12. mukushla
    • 2 years ago
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    is this correct?\[\tan^{-1}x +\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy} \]

  13. AravindG
    • 2 years ago
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    yep

  14. AravindG
    • 2 years ago
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    thats why i thought of an analogous for 3x :P

  15. AravindG
    • 2 years ago
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    i mean x,y ,z

  16. mukushla
    • 2 years ago
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    is this correct?\[\tan^{-1}x +\tan^{-1}y+\tan^{-1}z=\tan^{-1}\frac{x+y}{1-xy}+\tan^{-1}z\\=\tan^{-1}\frac{z+\frac{x+y}{1-xy}}{1-z\frac{x+y}{1-xy}}=\tan^{-1}\frac{x+y+z-xyz}{1-xy-xz-zy}\] so its valid

  17. mukushla
    • 2 years ago
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    lol.....ignore' is this correct?'

  18. experimentX
    • 2 years ago
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    lol .. that's correct!!

  19. AravindG
    • 2 years ago
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    :P thx a lot!!!!!!!!!!!!!!!!!!!111

  20. mukushla
    • 2 years ago
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    yw :)

  21. siddhantsharan
    • 2 years ago
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    Just remember that it is not valid for x belonging to R due to the domain range conditions you may need to add subtract pi. Otherwise its fine.

  22. siddhantsharan
    • 2 years ago
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    @AravindG

  23. AravindG
    • 2 years ago
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    k thx

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