If `AD,BE,CF` are internal bisectors of the angles of `DeltaABC` then `cos(A/2)/(AD)+cos(B/2)/(BE)+cos(C/2)/(CF)=`

If AD, BE , CF are internal bisectors of the angles of Delta ABC, " then " (cos A//2)/(AD) + (cos B//2)/(BE) + (cos C//2)/(CF) =

In a DeltaABC, cos""(A)/(2)cos""(B)/(2)cos""C/(2) is

In a DeltaABC, cos""(A)/(2)cos""(B)/(2)cos""C/(2) is

If alpha, beta, gamma are lenghts of internal bisectors of angles A,B,C respectively of DeltaABC , then sum (1)/(alpha) "cos" (A)/(2) is

In DeltaABC , prove that: a(cos^(2)C/2-cos^(2)B/2)=(b-c).cos^(2)A/2

AD, BE, CF are internal angular bisectors of Delta ABC and I is the incentre. If a(b+c)sec.(A)/(2)ID+b(a+c)sec.(B)/(2)IE+c(a+b)sec.(C )/(2)IF=kabc , then the value of k is (a) 1 (b) 2 (c) 3 (d) 4