In `DeltaABC,(b-c)/(b+c)cot""A/2+(b+c)/(b-c) tan""A/2=`

In any triangle , prove that ((b-c)/( b+c)) cot ""(A)/(2) +((b+c)/(b-c)) tan ""(A)/(2) = 2 cosec (B-C)

In any DeltaABC, sum (b - c) cot A/2 =

In any DeltaABC, sum (b - c) cot A/2 =

In DeltaABC,(a+b+c)(tan(A/2)+tan (B/2))=

In a DeltaABC,(a+b+c)(tan""A/2+tan""B/2)=

In a DeltaABC,sum(b+c)tan""A/2tan((B-C)/2)=

In a DeltaABC , if tan""(A)/(2)=(5)/(6),tan""(C )/(2)=(2)/(5) , then a)a, c, b are in AP b) a, b, c are in GP c)b, a, c are in AP d)a, b, c are in AP

In DeltaABC , tan""((A+B)/(2))*tan""(C)/(2)=

In DeltaABC,r(cot""A/2+cot""B/2+cot""C/2)=