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

1. DLS

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

2. DLS

Master @ParthKohli

3. ParthKohli

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

4. DLS

Hmm,nevermind tag someone better <3

5. ParthKohli

I have called the real master here.

6. DLS

I await to see him.

7. ParthKohli

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

8. ParthKohli

He says he's coming.

9. DLS

I await I await!

10. DLS

Lets see who it is D:

11. DLS

oooooo :")

12. ParthKohli

Told you.

13. mathslover

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

14. DLS

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

15. DLS

=)

16. mathslover

17. DLS

@mathslover I just substituted x=tan theta

18. mathslover

Ok!

19. sauravshakya

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

20. sauravshakya

Then use the property given by @mathslover

21. DLS

yo!

22. ParthKohli

I don't fit in this genius world.

23. DLS

but substitution method is better :P

24. sauravshakya

@ParthKohli may be because u r better than genius

25. DLS

^^

26. 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.

27. ParthKohli

I am overrated -_-