Show that the difference betwee f(x)=arcsin(x-1)/(square root(2(1+x^2))) and g(x)=arctg x is constant in (-1,inf) ?

### 2 Answers | Add Yours

`f(x)=arcsin((x-1)/(sqrt(2(1+x^2))))` and `g(x)=arctgx`

Differentiating f(x) with respect to x, applying chain rule and quotient rule,

`f’(x)=(sqrt(2(1+x^2))*1-((x-1)*sqrt2x)/sqrt(1+x^2))/(2(1+x^2)) *1/(sqrt(1-((x-1)^2)/(2(1+x^2))))`

`=(sqrt2((1+x^2)-(x^2-x)))/(2(1+x^2)sqrt(1+x^2)) *sqrt(2(1+x^2))/(sqrt(2+2x^2-x^2+2x-1)))`

`=((1+x)sqrt2sqrt2sqrt(1+x^2))/(2(1+x^2)sqrt(1+x^2)(x+1))`

`=1/(1+x^2)`

g(x)=arctgx

Differentiating g(x) with respect to x,

`g'(x)=1/(1+x^2)`

Since both derivatives are equal, the functions are the same, or their difference is a constant term.

Therefore, `f(x)-g(x)` =constant C

(where, `-1lt=xlt=oo` ).

Hence the proof.

**Sources:**

Let define a function

`h(x)=f(x)-g(x)` , `x in (-1,oo)`

`f(x)=sin^(-1)((x-1)/sqrt(2(1+x^2)))`

`g(x)=tan^(-1)(x)`

`` Let `tan^(-1)(x)=y`

`tan(y)=x` (i)

Thus

`g(x)=y` (ii)

`(x-1)/sqrt(2(1+x^2))=(tan(y)-1)/(sqrt(2(1+tan^2(y))))`

`=(tan(y)-1)/sqrt(2sec^2(y))`

`=(1/sqrt(2))(tan(y)-1)/sec(y)`

`=(1/sqrt(2))(sin(y)-cos(y))/(sec(y)cos(y))`

`=(1/sqrt(2))sin(y)-(1/sqrt(2))cos(y)`

`=sin(y)cos(pi/4)-cos(y)sin(pi/4)`

`=sin(y-pi/4)`

`f(x)=sin^(-1)((x-1)/sqrt(2(1+x^2)))=sin^(-1)(sin(y-pi/4))`

`=y-pi/4`

Thus

`h(x)=f(x)-g(x)=y-pi/4-y=-pi/4` ,which constant.

Hence proved.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes