A triangle `PQR , /_R=90^@` and `tan(P/2)` and `tan(Q/2)` roots of the `ax^2+bx+c=0` then prove that `a+b=c`

In a triangle PQR, ∠R=π/2 .If tan(P/2) & tan(Q/2) , are the roots of the equation ax^2+bx+c=(a≠0) then

In Delta PQR , /_R=pi/4 , tan(P/3) , tan(Q/3) are the roots of the equation ax^2+bx+c=0 , then

In any triangle PQR, angleR = pi/2 . If tan 'P/2 and tan 'Q/2 are the roots of the equation ax^(2) + bx + c=0 (a ne 0) , then show that, a+b=c.

In a triangle PQR angle R=(pi)/(4), if tan (p/3) and tan ((Q)/(3)) are the roots of the equation ax ^2 + bx +c=0 then

2x^(2) + 3x - alpha = 0 " has roots "-2 and beta " while the equation "x^(2) - 3mx + 2m^(2) = 0 " has both roots positive, where " alpha gt 0 and beta gt 0. Delta PQR, angleR=pi/2. " If than "(P/2) and tan (Q/2) " are the roots of the equation " ax^(2) + bx + c = 0 , then which one of the following is correct ?

If the tan theta and sec theta are roots of the equation ax^(2)+bx+c=0, then

If tan^(2)theta, sec^(2)theta are the roots of ax^(2)+bx+c=0 then b^(2)- a^(2) =

if ax^2+bx+c = 0 has imaginary roots and a+c lt b then prove that 4a+c lt 2b

In Delta ABC if tan(A/2) tan(B/2) + tan(B/2) tan(C/2) = 2/3 then a+c

If the roots of ax^(2) + bx + c = 0 are 2, (3)/(2) then (a+b+c)^(2)