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tan 3A-tan 2A-tan A= is equal to...

`tan 3A-tan 2A-tan A=` is equal to

A

`tan3Atan2AtanA`

B

`-tan3Atan2AtanA`

C

`tanAtan2A-tan2Atan3A-tan3AtanA`

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
A
Let `" "3A=A+2A`
`" "tan3A=tan(A+2A)`
`rArr" "tan3A=(tanA+tan2A)/(1-tanA*tan2A)`
`rArr" "tanA+tan2A=tan3A-tan3A*tan2A*tanA`
`rArr" "tan3A-tan2A-tanA=tan3A*tan2A*tanA`
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