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) ?

Asked on by ruals

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aruv | High School Teacher | (Level 2) Valedictorian

Posted on

Let define a function

`h(x)=f(x)-g(x)`  , `x in (-1,oo)`



`` Let  `tan^(-1)(x)=y`

`tan(y)=x`     (i)


`g(x)=y`        (ii)












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

Hence proved.

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llltkl | College Teacher | (Level 3) Valedictorian

Posted on

`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)))`




Differentiating g(x) with respect to x,


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.


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