Question

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

Answer 1

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

Answer 2

Master @ParthKohli

Answer 3

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

Answer 4

Hmm,nevermind tag someone better <3

Answer 5

I have called the real master here.

Answer 6

I await to see him.

Answer 7

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

Answer 8

He says he's coming.

Answer 9

I await I await!

Answer 10

Lets see who it is D:

Answer 11

oooooo :")

Answer 12

Told you.

Answer 13

Got to go , I am giving you a hint :

\[\tan ^{-1 } (\cfrac{\sqrt{1+x^2} - 1}{\sqrt{1+x^2-1}})\]

Answer 14

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

Answer 15

=)

Answer 16

Will this property help you?

Answer 17

@mathslover I just substituted x=tan theta

Answer 18

Ok!

Answer 19

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

Answer 20

Then use the property given by @mathslover

Answer 21

yo!

Answer 22

I don't fit in this genius world.

Answer 23

but substitution method is better :P

Answer 24

@ParthKohli may be because u r better than genius

Answer 25

^^

Answer 26

: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.

Answer 27

I am overrated -_-