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If (tan3A)/(tanA)=k, then (sin3A)/(sinA)...

If `(tan3A)/(tanA)=k,` then `(sin3A)/(sinA)=`

A

`(2k)/(k-1), k inR`

B

`(2k)/(k-1), k in[1//3,3]`

C

`(2k)/(k-1), k !in[1//3,3]`

D

`(k-1)/(2k), k !in[1//3,3]`

Text Solution

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The correct Answer is:
C
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