1.

If `tan(alpha+theta)=ntan(alpha-theta)`, show that `(n+1)sin2theta=(n-1)sin2alphadot`

Answer» `tan(alpha+theta) = n tan(alpha-theta)`
`=>tan(alpha+theta)/(tan(alpha-theta)) = n/1`
Using componendo and dividendo,
`=>(tan(alpha+theta)+tan(alpha-theta))/(tan(alpha+theta)-tan(alpha-theta)) = (n+1)/(n-1)`
`=>(sin(alpha+theta)/cos(alpha+theta)+sin(alpha-theta)/cos(alpha-theta))/(sin(alpha+theta)/cos(alpha+theta)-sin(alpha-theta)/cos(alpha-theta)) = (n+1)/(n-1)`
`=>(sin(alpha+theta)cos(alpha-theta)+cos(alpha+theta)sin(alpha-theta))/(sin(alpha+theta)cos(alpha-theta)+cos(alpha+theta)sin(alpha-theta)) =(n+1)/(n-1)`
`=>(sin(alpha+theta+alpha-theta))/(sin(alpha+theta-alpha+theta))=(n+1)/(n-1)`
`=>(sin2alpha)/(sin2theta) = (n+1)/(n-1)`
`=>(n-1)sin2alpha = (n+1)sin2theta.`


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