In any ΔABC,1r1+1r2+1r3 is equal to
3/r
1/r
2/r
1
We have, 1r1+1r2+1r3=s−aΔ+s−bΔ+s−cΔ=3s−(a+b+c)Δ=3s−2sΔ=sΔ=1r
In any ΔABC, 4(sa−1)(sb−1)(sc−1) is equal to
If P(Q−r)x2+Q(r−P)x+r(P−Q)=0 has equal roots then 2Q=(where P,Q,r ϵ R)