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If cos2alpha=(3cos2beta-1)/(3-cos2beta)...

If `cos2alpha=(3cos2beta-1)/(3-cos2beta)`, then tan `alpha` is equal to

A

`sqrt(2)tanbeta`

B

`tanbeta`

C

`sin2 beta`

D

`sqrt(2)cotbeta`

Text Solution

Verified by Experts

The correct Answer is:
A
Given ,`cos2alpha=(3cos2beta-1)/(3-cos2beta)`
`rArr(1-tan^(2)alpha)/(1+tan^(2)alpha)=(3((1-(tan^(2)beta))/(1+tan^(2)beta))-1)/(3-((tan^(2)beta)/(1+tan^(2)beta)))`
`rArr(1-tan^(2)alpha)/(1+tan^(2)alpha)=(2-4tan^(2)beta)/(2+4tan^(2)beta)=(1-2tan^(2)beta)/(1+2tan^(2)beta)`
Applying componendo and dividendo ,we get
`(1)/(tan^(2)alpha)=(1)/(2tan^(2)beta)rArrtanalpha=sqrt(2)tanbeta`
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