Question

Write it in SIMPLEST FORM.

Answer 1

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

Answer 2

My Approach :-
Put \[\large \bf \tan^{-1} x=\theta\]
\[\large \bf x=\tan \theta\]
plug in the question,
\[\large \bf \tan^{-1} (\sqrt{1+\tan^2 \theta}+\tan \theta)\]
\[\large \bf \tan^{-1} (\sec \theta + \tan \theta)\]

Answer 3

then,
\[\large \bf \tan^{-1} (\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta})\]
\[\large \bf \tan^{-1} ( \frac{1+\sin \theta}{\cos \theta})\]

Answer 4

What to do next ???

Answer 5

i guess you cannot use calculus right?

Answer 6

what do you mean ?

Answer 7

@satellite73

Answer 8

i am sure there is some algebra gimmick for this, but i do not see it at the moment
if you can use calculus then it is not too hard at all

Answer 9

so what next ?

Answer 10

if you start here

\[ \tan^{-1} (\sec \theta + \tan \theta)\]
take the derivative

Answer 11

why we take derivative of this ?
We have to simplify it only ALGEBRAICALLY !
I mean i have to find the simplest form in terms of x

Answer 12

i think we have to solve it by ALGEBRAICALLY

Answer 13

yeah that is what i thought, so forget this method

Answer 14

yup !

Answer 15

if i convert in half angle formula, then no term will cancel out !

Answer 16

\[\huge \bf \tan^{-1} (\frac{1+2\sin \frac{\theta}{2} \cos \frac{\theta}{2}}{1-2\sin^2 \frac{\theta}{2}})\]

Answer 17

@satellite73

Answer 18

@iGreen

Answer 19

Let:\[\theta=\arctan(x+\sqrt{1+x^2})\]\[\therefore\tan(\theta)=x+\sqrt{1+x^2}\]\[\therefore\tan^2(\theta)=x^2+2x\sqrt{1+x^2}+1+x^2=1+2x^2+2x\sqrt{1+x^2}\]\[\therefore\tan(2\theta)=\frac{2\tan(\theta)}{1-\tan^2(\theta)}\]\[=\frac{2(x+\sqrt{1+x^2})}{-2x^2-2x\sqrt{1+x^2}}=\frac{2(x+\sqrt{1+x^2})}{-2x(x+\sqrt{1+x^2})}=-\frac{1}{x}\]\[\therefore2\theta=\arctan(-\frac{1}{x})=-\arctan(\frac{1}{x})\]\[\therefore\theta=-\frac{1}{2}\arctan(\frac{1}{x})\]\[\therefore\arctan(x+\sqrt{1+x^2})=-\frac{1}{2}\arctan(\frac{1}{x})\]

Answer 20

I'm not sure if it can be simplified any further

Answer 21

i have another way !

Answer 22

i like your method !

Answer 23

Another Method :-
Let \[\large \bf x=\cot \theta\]
\[\large \bf \tan^{-1} (\sqrt{1+\cot^2 \theta}+\cot \theta)\]
\[\large \bf \tan^{-1} (cosec \theta + \cot \theta)\]
\[\large \bf \tan^{-1} (\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta})\]
\[\large \bf \tan^{-1} (\frac{1+\cos \theta}{\sin \theta})\]
\[\large \bf \tan^{-1} (\cot \frac{\theta}{2})\]
\[\large \bf \tan^{-1} (\tan(\frac {\pi}{2}- \frac{\theta}{2}))\]
\[\large \bf \frac{\pi}{2}-\frac{\theta}{2}\]
\[\large \bf \frac{\pi}{2}-\frac{\cot^{-1}}{x}\]

Answer 24

@asnaseer

Answer 25

Am i right ?

Answer 26

Looks right :)

Answer 27

thanks :)