tan A+tan B+tan C-(sin(A+B+C))/(cos A cos B cos C)=

If,A,B,C are angles of a triangle and if none of them is equal to (pi)/(2) ,then tan A+tan B+tan C is cot A cotB cot C tan A tan B tan C sin A sin B sin C cos A cos B cos C

In a Delta ABC , If tan A/2, tan B/2, tan C/2 are in A.P. then cos A. cos B, cos C are in

In a Delta ABC , If tan A/2, tan B/2, tan C/2 are in A.P. then cos A. cos B, cos C are in

If A+B+C =pi , sin A sin B sin C=p and cos A cos B cos C=q then obtain the cubic equation whose roots are tan A, tan B and tan C.

In a Delta ABC , If tan\ A/2, tan\ B/2, tan\ C/2 are in A.P. then cos A, cos B, cos C are in

If (x)/(y)=(cos A)/(cos B) where A!=B ,then (a) tan(A+B)/(2)=(x tan A+y tan B)/(x+y) (b) tan(A-B)/(2)=(x tan A-y tan B)/(x+y) (c) (sin(A+B))/(sin(A-B))=(y sin A+x sin B)/(y sin A-x sin B) (d) (x cos B-y cos A)/(xcosB+ ycosA)=0

If A + B + C = 180^(@) prove that: (i) "sin" (B+C-A) + "sin"(C+ A-B) + sin (A + B-C) = 4 sin A sin B sin C (ii) cos(- A+B+C) + cos (A - B + C) + cos (A + B - C) = 1 + 4 cos A cos B cos C (iii) "tan" (B+C - A) + tan (C+A-B) + tan (A + B-C) = tan (B+C - A) tan (C+ A - B) tan (A+B-C)

If A + B + C = pi " and " cos A = cos B cos C, "show that" tan A = tan B + tan C