If in a `!ABC` , tan `A/2` and tan `B/2` are the roots of the equation `6x^(2)-5x+1=0`, then

Options

A. `a^(2)+b^(2)gtc^(2)`
B. `a^(2)-b^(2)=c^(2)`
C. `a^(2)+b^(2)=c^(2)`
D. none of these

Explanation

We have,
`tanA/2+tanB/2=5/6` and `tanA/2tanB/2=1/6`.
`thereforetan(A/2+B/2)=(tanA/2+tanB/2)/(1-tanA/2tanB/2)`
`rArrtan(A/2)+B/2)=(5/6)/(1-1/6)=1`
`rArrA/2+B/2=pi/4rArrA+B=pi/2rArrangleC=pi/2rArrc^(2)=a^(2)+b^(2)`