" (ii) "tan(B+C)/(2)=cot(A)/(2)

tan((A+B)/(2))="cot"(C)/(2)

If A,B,C are the three angles of a triangle,show that:(iv) tan((B+C)/2)=cot(A/2)

If A,B,C are angles of a triangle, prove that "tan "(B+C)/(2)="cot"(A)/(2) .

Consider the following for triangle ABC : 1. sin((B+C)/(2))=cos((A)/(2)) 2. tan((B+C)/(2))=cot((A)/(2)) 3. sin (B + C) = cos A 4. tan(B + C) = -cot A Which of the above are correct ?

If A,B,C are the interior angles of a triangle ABC, prove that (tan(B+C))/(2)=(cot A)/(2)

If tan (B-C)/(2)=x cot (A)/(2) then x=

In DeltaABC show that (a+b+c)(tan.(A)/(2)+tan.(B)/(2))=2cot.(C)/(2)

In a triangle ABC, prove that (a)cos(A+B)+cos C=0(b)tan((A+B)/(2))=cot((C)/(2))

In triangle ABC, prove that tan(B+C)/2=cot(A/2)

In a triangle ABC, prove that (a) cos(A + B) + cos C = 0 (b) tan( (A+B)/2)= cot(C/ 2)