DLS
  • DLS
Simplify this! \large \tan^{-1} \frac{ \sqrt{1+x^2}-1}{x},x \neq 0
Mathematics
chestercat
  • chestercat
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DLS
  • DLS
\[\large \tan^{-1} \frac{ \sqrt{1+x^2}-1}{x},x \neq 0\]
DLS
  • DLS
ParthKohli
  • ParthKohli
Thanks for calling me, but I suck at trig :-P

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DLS
  • DLS
Hmm,nevermind tag someone better <3
ParthKohli
  • ParthKohli
I have called the real master here.
DLS
  • DLS
I await to see him.
ParthKohli
  • ParthKohli
He has seen my FB message. Let's see when he's here.
ParthKohli
  • ParthKohli
He says he's coming.
DLS
  • DLS
I await I await!
DLS
  • DLS
Lets see who it is D:
DLS
  • DLS
oooooo :")
ParthKohli
  • ParthKohli
Told you.
mathslover
  • mathslover
Got to go , I am giving you a hint : \[\tan ^{-1 } (\cfrac{\sqrt{1+x^2} - 1}{\sqrt{1+x^2-1}})\]
DLS
  • DLS
\[\LARGE \tan^{-1} \frac{\sec \theta-1}{\tan \theta}=> \frac{1-\cos \theta}{\sin \theta}\] used half angle formula and got it :
DLS
  • DLS
=)
mathslover
  • mathslover
Will this property help you?
DLS
  • DLS
@mathslover I just substituted x=tan theta
mathslover
  • mathslover
Ok!
anonymous
  • anonymous
multiply both numerator and denominator by root(1+x^2) +1
anonymous
  • anonymous
Then use the property given by @mathslover
DLS
  • DLS
yo!
ParthKohli
  • ParthKohli
I don't fit in this genius world.
DLS
  • DLS
but substitution method is better :P
anonymous
  • anonymous
@ParthKohli may be because u r better than genius
DLS
  • DLS
^^
mathslover
  • mathslover
: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.
ParthKohli
  • ParthKohli
I am overrated -_-

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