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

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DLS
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large \tan^{1} \frac{ \sqrt{1+x^2}1}{x},x \neq 0\]

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks for calling me, but I suck at trig :P

DLS
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm,nevermind tag someone better <3

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0I have called the real master here.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0He has seen my FB message. Let's see when he's here.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0He says he's coming.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.1Got to go , I am giving you a hint : \[\tan ^{1 } (\cfrac{\sqrt{1+x^2}  1}{\sqrt{1+x^21}})\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.0\[\LARGE \tan^{1} \frac{\sec \theta1}{\tan \theta}=> \frac{1\cos \theta}{\sin \theta}\] used half angle formula and got it :

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.1Will this property help you?

DLS
 3 years ago
Best ResponseYou've already chosen the best response.0@mathslover I just substituted x=tan theta

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0multiply both numerator and denominator by root(1+x^2) +1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then use the property given by @mathslover

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0I don't fit in this genius world.

DLS
 3 years ago
Best ResponseYou've already chosen the best response.0but substitution method is better :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ParthKohli may be because u r better than genius

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.1: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.
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