A + C = B then tan A tan B tan C=

Given that A = B +C. prove that tan A - tan B - tan C = tan A tan B tan C.

Statement-1: In any DeltaABC if A is obtuse, then tanBtanC lt1 Statement-2: In any !ABC , we have tan A + tan B + tan C = tan A tan B tan C

If Delta ABC , if tan A + tan B + tan C = 3sqrt(3) , then the triangle is :

If A+ B+ C= pi , n in Z,then tan nA+ tan +nB +tan nC is equal to

If cos(A+B) sin (C-D) = cos (A-B) sin (C+D) , then show that tan A tan B tan C + tan D =0

If A +B+C= pi and /_C is obtuse then tan A. tan B is

In a triangle tan A+ tan B + tan C=6 and tan A tan B= 2, then the values of tan A, tan B and tan C are

If A, B, C are angles of Delta ABC and tan A tan C = 3, tan B tan C = 6 , then

(cos(A-B))/(cos(A+B))+cos(C+D)/(cos(C-D))=0 => tan A tan B tan C tan D =

Show that the orthocentre of DeltaABC having vertices A(x_1, y_1), B(x_2, y_2), C(x_3, y_3) is : (x_1 tan A + x_2 tan B + x_3 tan C)/(tan A + tan B + tan C) , (y_1 tan A + y_2 tan B + y_3 tan C)/(tan A + tan B + tan C))