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



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DLS
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\[\large \tan^{1} \frac{ \sqrt{1+x^2}1}{x},x \neq 0\]

DLS
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Master @ParthKohli

ParthKohli
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Thanks for calling me, but I suck at trig :P

DLS
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Hmm,nevermind tag someone better <3

ParthKohli
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I have called the real master here.

DLS
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I await to see him.

ParthKohli
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He has seen my FB message. Let's see when he's here.

ParthKohli
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He says he's coming.

DLS
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I await I await!

DLS
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Lets see who it is D:

DLS
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oooooo :")

ParthKohli
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Told you.

mathslover
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Got to go , I am giving you a hint :
\[\tan ^{1 } (\cfrac{\sqrt{1+x^2}  1}{\sqrt{1+x^21}})\]

DLS
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\[\LARGE \tan^{1} \frac{\sec \theta1}{\tan \theta}=> \frac{1\cos \theta}{\sin \theta}\]
used half angle formula and got it :

DLS
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=)

mathslover
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Will this property help you?

DLS
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@mathslover I just substituted x=tan theta

mathslover
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Ok!

sauravshakya
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multiply both numerator and denominator by root(1+x^2) +1

sauravshakya
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Then use the property given by @mathslover

DLS
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yo!

ParthKohli
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I don't fit in this genius world.

DLS
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but substitution method is better :P

sauravshakya
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@ParthKohli may be because u r better than genius

DLS
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^^

mathslover
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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.

ParthKohli
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I am overrated _