tan 3A − tan 2A − tan A is Equal To

Solution

Using the identity:

\[ \tan(X+Y) = \frac{\tan X+\tan Y} {1-\tan X\tan Y} \]

Since

\[ 3A=2A+A \]

we have

\[ \tan3A = \frac{\tan2A+\tan A} {1-\tan2A\tan A} \]

Cross multiplying,

\[ \tan3A(1-\tan2A\tan A) = \tan2A+\tan A \]

Expanding,

\[ \tan3A – \tan3A\tan2A\tan A = \tan2A+\tan A \]

Rearranging,

\[ \tan3A-\tan2A-\tan A = \tan3A\tan2A\tan A \]

Final Answer

\[ \boxed{ \tan3A-\tan2A-\tan A = \tan3A\tan2A\tan A } \]

Correct Option: (a)