UPSEE Solved Model Paper 6

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) = $\frac{\tan A + \tan B}{1 - \tan A \tan B}$ 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 = $\frac{\tan 2A + \tan A}{1 - \tan 2A \tan A}$ Since tan (A + B) = $\frac{\tan 2A + \tan A}{1 - \tan A \tan B}$ 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