(v)sec(A+B-C)/(2)=cosec C

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

If a,b,c are in A.P. Then cos "" ( A-C)/( 2) cosec ""(B)/(2) =

In DeltaABC , prove that "cosec"(A)/(2)+"cosec"(B)/(2)+"cosec"(C)/(2) ge 6 .

In DeltaABC , prove that "cosec"(A)/(2)+"cosec"(B)/(2)+"cosec"(C)/(2) ge 6 .

In DeltaABC , prove that "cosec"(A)/(2)+"cosec"(B)/(2)+"cosec"(C)/(2) ge 6 .

In DeltaABC , prove that "cosec"(A)/(2)+"cosec"(B)/(2)+"cosec"(C)/(2) ge 6 .

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

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

The angle A of the triangle ABC is obtuse, if sec(B+C)=cosec (B-C)=2,find the angles.

If (sin^(-1)a)^(2)+(cos^(-1)b)^(2)+(sec^(-1)c)^(2)+(cosec^(-1)d)^(2)=(5pi)^(2)/2 , then the value of (sin^(-1)a)^(2)-(cos^(-1)b)^(2)+(sec^(-1)c)^(2)-(cosec^(-1)d)^(2) is equal to