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If alphaa n dbeta are acute angles satis...

If `alphaa n dbeta` are acute angles satisfying `cos2alpha=(3cos2beta-1)/(3-cos2beta),then
tanalpha=` `sqrt(2)tanbeta` (b) `1/(sqrt(2))t a nbeta` (c) `sqrt(2)cotbeta` (d) `1/(sqrt(2))cotbeta`

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`cos2alpha=(3cos2beta-1)/(3-cos2beta)`
`=>(cos2alpha-1)/(cos2alpha+1)=((3cos2beta-1)-(3-cos2beta))/((3cos2beta-1+(3-cos2beta))`
`=>(cos2alpha-1)/(cos2alpha+1)=(4cos2beta-4)/(2cos2beta+2)`
Taking `(-1)` common from L.H.S side and `(-4)`
common from R.H.S side we have,
`=>(-1(1-cos2alpha))/(1+cos2alpha)=(-4(1-cos2beta))/(2(1+cos2beta))`
`=>(1-cos2alpha)/(1+cos2alpha)=2(1-cos2beta)/(1+cos2beta)`
`=>(2sin^2alpha)/(2cos^2alpha)=2(2sin^2beta)/(2cos^2beta)`
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