In △ABC,1−tan A2 tan B2= [Roorkee 1973]
2ca+b+c
aa+b+c
2a+b+c
4aa+b+c
1−tanA2 tan B2=cosA2cosB2−sinA2sinB2sinB2cosA2cosB2 =cos(A2+B2)cosA2cosB2=sinC2cosA2cosB2 =[(s−a)(s−b)bc.acab.s(s−a)s(s−b)]1/2=cs=2ca+b+c.
If in a triangle ABC, ∠C=60o, then 1a+c+1b+c= [IIT 1975]
In △ABC,1−tan A2 tan B2= [Roorkee 1973]