If `a , b , c >0` and `s=(a+b+c)/2,p rov et h a t` `tan^(-1)sqrt((2a s)/(b c))+tan^(-1)sqrt((2b s)/(c a))+tan^(-1)sqrt((2c s)/(a b))=pi`

Explanation

L.H.S=`tan^(-1)sqrt((2a s)/(b c))+tan^(-1)sqrt((2b s)/(c a))+tan^(-1)sqrt((2c s)/(a b))`

So,`tanalpha=sqrt((2a s)/(b c))=sqrt((a (a+b+c))/(b c))`

`tanbeta=sqrt((2b s)/(c a))=sqrt((b (a+b+c))/(c a))`

`tangamma=sqrt((2c s)/(a b))=sqrt((c (a+b+c))/(a b))`

Thus,`tanalpha+tanbeta+tangamma=sqrt((a (a+b+c))/(b c))+sqrt((b (a+b+c))/(c a))+sqrt((c (a+b+c))/(a b))`

=`​(a+b+c)^(3/2)/sqrt(abc)`

=`tanαtanβtanγ`

or `tantheta=0`

...