Use the method from this example $to$ prove the identity.

$arctan(x)+arctan(\frac{1}{x})=\frac{}{2},x>0$

Let $f(x)=arctan(x)+arctan(\frac{1}{x}).$ Then

$on$ the given domain, $(0,).$ Therefore, $f(x)=C$ for some constant $C0,f^{\text{'}}(x)=0$ for all $xinan$ interval $(a,b),$ then $fis$ constant $on(a,b).$

$To$ find $C,we$ let $x=1to$ get

$C=f(1)=arctan(1)+arctan(.,1)C$

$=>\frac{}{4}+,,-=\frac{}{2}$

Thus, $f(x)=\frac{}{2},$ that $is,arctan(x)+arctan(\frac{1}{x})=\frac{}{2}.$