If `cos theta=a/(b+c), cos phi= b/(a+c)` and `cos psi=c/(a+b)` where `theta, phi, psi in (0, pi)` and `a,b,c` are sides of triangle `ABC` then `tan^2(theta/2)+tan^2(phi/2)+tan^2(psi/2)=`

Answer

1

Explanation

`tan^(2)""(theta)/(2)+tan^(2)""(phi)/(2)+tan^(2)""(psi)/(2)`
`=(1-cos theta)/(1+cos theta)+(1-cos phi)/(1+cos phi)+(1-cos psi)/(1+cos psi)`
`=(1-(a)/(b+c))+(1-(b)/(c+a))/(1+(b)/(c+a))+(1-(c)/(a+b))/(1+(c)/(a+b))`
`=(b+c-a)/(a+b+c)+(c+a-b)/(a+b+c)+(a+b-c)/(a+b+c)`
`=(a+b+c)/(a+b+c)`