1.

If `3sinbeta=sin(2alpha+beta)`then `tan(alpha+beta)-2tanalpha`is(a)independent of `alpha`(b)independent of `beta`(c)dependent of both `alpha` and `beta`(d)independent of both `alpha` and `beta`

Answer» `tan(alpha+beta) - 2tanalpha`
`= sin(alpha+beta)/cos(alpha+beta) - sinalpha/cosalpha - tanalpha`
`=(sin(alpha+beta)cosalpha - sinalphacos(alpha+beta))/(cosalphacos(alpha+beta)) - tanalpha`
`=(sin(alpha+beta-alpha))/(cosalphacos(alpha+beta)) - sinalpha/cosalpha`
`=(sinbeta)/(cosalphacos(alpha+beta)) - sinalpha/cosalpha`
`=(sinbeta - sinalphacos(alpha+beta))/(cosalphacos(alpha+beta)) `
`=(2(sinbeta - sinalphacos(alpha+beta)))/(2cosalphacos(alpha+beta)) `
`=(2sinbeta - 2sinalphacos(alpha+beta))/(2cosalphacos(alpha+beta)) `
`=(2sinbeta - (sin(alpha+alpha+beta) + sin(alpha-alpha-beta)))/(2cosalphacos(alpha+beta))`
`=(2sinbeta - (sin(2alpha+beta) - sinbeta))/(2cosalphacos(alpha+beta))`
`=(3sinbeta - sin(2alpha+beta))/(2cosalphacos(alpha+beta))`
It is given that, `3sinbeta = sin(2alpha+beta) => 3sinbeta - sin(2alpha+beta) = 0`
So, it becomes,
`=0/(2cosalphacos(alpha+beta))`
`=0`
`:. tan(alpha+beta) - 2tanalpha =0`
So, given expression is independent of both `alpha` and `beta`.


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