किसी त्रिभुज ABC में सिद्ध कीजिए कि-
`(a+b+c)/(a+b-c)=cot""(A)/(2)cot""(B)/(2)`

In Delta ABC, cot ""(A)/(2) +cot""(B)/(2) + cot""(C)/(2)=

a^(2)cot A+b^(2)cot B+c^(2)cot C=

In Delta ABC, " if" s= 2b " then " cot "" (A)/(2) cot "" ( C)/(2) =

In /_ABC, if 3a=b+c, then cot((B)/(2))cot((C)/(2))=

In /_\ ABC, if 2b=a+c then cot(A/2)cot(C/2)=

In any Delta ABC, porve that (a+b-c)cot((B)/(2))=(a-b+c)cot((C)/(2))

In Delta ABC, if s=2b then cot""A/2 cot ""C/2 =

In Delta ABC , if a=3 , b=4,c=6 then ("cot" (A)/(2)+"cot" (B)/(2)+ "cot" ( C)/(2))/("cot" A + "cot" B + "cot" C)=