Let `a , ba n dc` be positive real numbers. `l e ttheta=tan^(-1)sqrt((a(a+b+c))/(b c))+tan^(-1)sqrt((b(a+b+c))/(c a))+tan^(-1)sqrt((c(a+b+c))/(a b))dot` Then `tantheta=` __________

Explanation

Let
`S = tan^(-1) sqrt((a(a + b + c))/(bc)) tan^(-1) sqrt((b(a + b + c))/(ca)) + tan^(-1) sqrt((c (a + b + c))/(ab))`
Now, `sqrt((a(a + b + c))/(bc)) sqrt((b (a + b + c))/(ca)) = (a + b + c)/(c) = 1 + (b)/(c) + (a)/(c) gt 1`
`S = pi + tan^(-1). (sqrt((a (a + b + c))/(bc)) + sqrt((b(a + b + c))/(ca)))/(1- sqrt((a(a + b + c))/(bc)) sqrt((b(a + b + c))/(ca))) + tan^(-1) sqrt((c(a + b + c))/(ab))`
`= pi + tan^(-1). (sqrt((a + b + c)/(c)) (sqrt((a)/(b)) + sqrt((b)/(a))))/(1 - (a + b + c)/(c)) + tan^(-1) sqrt((c(a + b + c))/(ab))`
`= pi + tan^(-1) (- sqrt((c (a + b + c))/(ab)) ) + tan^(-1) sqrt((c(a + b + c))/(ab))`
`= pi`