Question
Mathematics Question on Trigonometric Functions
The value of tan3A−tan2A−tanA is equal to
A
tan3Atan2AtanA
B
−tan3Atan2AtanA
C
tanAtan2A−tan2Atan3A−tan3AtanA
D
None of these
Answer
tan3Atan2AtanA
Explanation
Solution
tan3A=tan(2A+A) ⇒tan3A=1−tan2AtanAtan2A+tanA ⇒tan3A−tan3Atan2AtanA =tan2A+tanA ⇒tan3A−tan2A−tanA =tan3Atan2AtanA