In a triangle `A B C ,/_C=pi/2dot` If `tan(A/2)` and `tan(B/2)` are the roots of the equation `a x^2+b x+c=0,(a!=0),` then the value of `(a+b)/c` (where `a , b , c ,` are sides of `` opposite to angles `A , B , C ,` respectively) is

Explanation

As `tan(A/2) and tan(B/2)` are the roots of `ax^2+bx+c = 0`

`:. tan(A/2)+tan(B/2) = -b/a`

`tan(A/2)tan(B/2) = c/a`

Now, it is given that,

`/_C = pi/2 and A+B+C = pi`

`:. A+B = pi-pi/2 = pi/2`

`=>(A+B)/2 = pi/4`

Now, `tan((A+B)/2) = tan (pi/4)`

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