[" घ) "/_ABC" में स्पिद्ध कीजिए "],[qquad sec((B+C)/(2))=cosec(A)/(2)]

For triangle ABC prove that sec((A+B)/2)="cosec"C/2

If I is the incenter of a triangle ABC, then the ratio IA:IB:IC is equal to cosec (A)/(2):csc(B)/(2):csc(C)/(2)sin(A)/(2):sin(B)/(2):sin(C)/(2)sec(A)/(2):sec(B)/(2):sec(C)/(2) none of these

In Delta ABC , if a,b,c are in A.P, then cos((A-C)/2)cosec(B/2)=

If A, B, C are interior angles of DeltaABC , the prove that cosec((A+B)/2)=sec(C/2) .

In a Delta ABC , the minimum value of sec^(2). (A)/(2)+sec^(2). (B)/(2)+sec^(2). (C)/(2) is equal to

In a Delta ABC , the minimum value of sec^(2). (A)/(2)+sec^(2). (B)/(2)+sec^(2). (C)/(2) is equal to

In a Delta ABC , the minimum value of sec^(2). (A)/(2)+sec^(2). (B)/(2)+sec^(2). (C)/(2) is equal to

int sec^(2)x*cosec^2(x)dx=.....+c

In a triangle ABC ,show that s.sec((A)/(2))sec((B)/(2))sec((C)/(2))=2[(abc)/(sin A sin B sin C)]^((1)/(3))

In hat harr ABC, prove that csc(A)/(2)+csc(B)/(2)+csc(C)/(2)>=6