Tan(A±B), Cot(A±B) Formulae

tan(A+B)=p,tan(A-B)=qthen cot2B=

Prove that: (tan(A+B))/(cot(A-B))=(tan^(2)A-tan^(2)B)/(1-tan^(2)A tan^(2)B)

In a DeltaABC, tan ""1/2 (A+B) . Cot""1/2(A-B) is equal to

(sin^(2)A-sin^(2)B)/(sin A cos A-sin B cos B) is equal to (a) sin A cos A-sin B cos Btan(A-B)(b)tan(A+B)cot(A-B)(d)cot(A+B)

In triangle ABC, if |[1,1,1], [cot (A/2), cot(B/2), cot(C/2)], [tan(B/2)+tan(C/2), tan(C/2)+ tan(A/2), tan(A/2)+ tan(B/2)]| then the triangle must be (A) Equilateral (B) Isoceless (C) Right Angle (D) none of these

In triangle ABC, if 1,1cot(A)/(2),cot(B)/(2)tan(B)/(2)+tan(C)/(2)quad tan(B)/(2)+tan(A)/(2)quad tan(A)/(2)+tan(B)/(2) obtuse angled d.none of these

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

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

If tan A - tan B = x and cot B - cot A =y , then what is cot ( A -B) equal to ?

If tan A=cot B, then A+B is equal to