We can solve this problem using the trigonometric function of sum and difference compound angles : An algebraic sum of two or more angles is called a compound angle tan (A + B) =
tanA+tanB
1−tanAtanB
By using this formula we can find the calue of tan 3A - tan 2A - tan A We have tan 3A - tan 2A - tan A Write , tan 3A as tan (2a + A) tan 3A = tan (2A + A) tan 3A =
tan2A+tanA
1−tan2AtanA
Since tan (A + B) =
tan2A+tanA
1−tanAtanB
tan 3A (1 - tan 2A tan A) = tan 2A + tan A tan 3A - tan 3A tan 2A tan A = tan 2A + tan A tan 3A - tan 2A - tan A = tan 3A . tan 2A . tan A Therefore tan 3A - tan 2A - tan A = tan 3A . tan 2A . tan A