`(a+b)/(a-b)=tan.(A+B)/(2)cot.(A-B)/(2)`

Prove that: (sin A+sin B)/(sin A-sin B)=tan((A+B)/2)cot((A-B)/2)

If (x)/(y)=(cos A)/(cos B)then(x tan A+y tan B)/(x+y)=cot(A+B)/(2)(b)cot(A-B)/(2)tan(A-B)/(2) (d) tan(A+B)/(2)

Prove that: (s in A+s in B)/(s in A-s in B)=tan((A+B)/(2))cot((A-B)/(2))

If A+B+C=pi, show that : tan.(A)/(2)tan.(B)/(2)+tan.(B)/(2)tan.(C)/(2)+tan.(C)/(2)tan.(A)/(2)=1 Hence deduce that : cot.(A)/(2)+cot.(B)/(2)+.cot.(C)/(2)=cot.(A)/(2).cot.(B)/(2)tan.(C)/(2) .

(sinA-sinB)/(sinA+sinB)=tan((A-B)/2).cot((A+B)/2)

(sinA-sinB)/(sinA+sinB)=tan((A-B)/2).cot((A+B)/2)

In triangle ABC,, prove that (tan(B-C))/(2),=[(b-c)/(b+c)](cot A)/(2)(tan(C-A))/(2),=[(c-a)/(c+a)](cot B)/(2)(tan(A-B))/(2),=[(a-b)/(a+b)](cot B)/(2)

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

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

If A+B+C=pi , prove that (a) tan 3A+tan3B+tan3C=tan3A tan 3B tan 3C (b) cot""(A)/(2)+cot""(B)/(2)+cot""(C)/(2)=cot""(A)/(2)cot""(B)/(2)cot""(C)/(2)