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DLS

  • one year ago

Simplify this! \large \tan^{-1} \frac{ \sqrt{1+x^2}-1}{x},x \neq 0

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  1. DLS
    • one year ago
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    \[\large \tan^{-1} \frac{ \sqrt{1+x^2}-1}{x},x \neq 0\]

  2. DLS
    • one year ago
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    Master @ParthKohli

  3. ParthKohli
    • one year ago
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    Thanks for calling me, but I suck at trig :-P

  4. DLS
    • one year ago
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    Hmm,nevermind tag someone better <3

  5. ParthKohli
    • one year ago
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    I have called the real master here.

  6. DLS
    • one year ago
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    I await to see him.

  7. ParthKohli
    • one year ago
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    He has seen my FB message. Let's see when he's here.

  8. ParthKohli
    • one year ago
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    He says he's coming.

  9. DLS
    • one year ago
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    I await I await!

  10. DLS
    • one year ago
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    Lets see who it is D:

  11. DLS
    • one year ago
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    oooooo :")

  12. ParthKohli
    • one year ago
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    Told you.

  13. mathslover
    • one year ago
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    Got to go , I am giving you a hint : \[\tan ^{-1 } (\cfrac{\sqrt{1+x^2} - 1}{\sqrt{1+x^2-1}})\]

  14. DLS
    • one year ago
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    \[\LARGE \tan^{-1} \frac{\sec \theta-1}{\tan \theta}=> \frac{1-\cos \theta}{\sin \theta}\] used half angle formula and got it :

  15. DLS
    • one year ago
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    =)

  16. mathslover
    • one year ago
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    Will this property help you?

  17. DLS
    • one year ago
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    @mathslover I just substituted x=tan theta

  18. mathslover
    • one year ago
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    Ok!

  19. sauravshakya
    • one year ago
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    multiply both numerator and denominator by root(1+x^2) +1

  20. sauravshakya
    • one year ago
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    Then use the property given by @mathslover

  21. DLS
    • one year ago
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    yo!

  22. ParthKohli
    • one year ago
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    I don't fit in this genius world.

  23. DLS
    • one year ago
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    but substitution method is better :P

  24. sauravshakya
    • one year ago
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    @ParthKohli may be because u r better than genius

  25. DLS
    • one year ago
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    ^^

  26. mathslover
    • one year ago
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    :P if you love substitution or you think it is better or simpler then go for it but do try the another method also. It would be beneficial for you.

  27. ParthKohli
    • one year ago
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    I am overrated -_-

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