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Correct Answer – D
We have `4x^(2)-16x+15lt0`
`rArr (3)/(2)ltxlt(5)/(2)`
Therefore, the integral solution of
`4x^(2)-16x+15lt0` is `x=2`.
Thus, `tan alpha=2`. It is given that `cos beta=tan45^(@)=1`.
`therefore sin(alpha+beta)sin(alpha-beta)=sin^(2)alpha-sin^(2)beta`.
`(1)/(1+cot^(2)alpha)(1-cos^(2)beta)`
`=(1)/(1+(1)/(4))-0=(4)/(5)`
It is given that slope of bisector is`cos beta.`
`:. cos beta = tan45^@ =>cos beta = 1->(1)`
Now, `4x^2-16x+15 lt 0`
`=>4x^2-10x-6x+15 lt 0`
`=>2x(2x-5)-3(2x-5) lt 0`
`=>(2x-3)(2x-5) lt 0`
`:. 3/2 lt x lt 5`
As, `tan alpha` is an integral solution for this inequality,
`:. tanalpha = 2`
`=>cotalpha = 1/2->(2)`
Now, `sin(alpha+beta)sin(alpha-beta) = sin^2alpha-sin^2beta`
`=1/(cosec^2alpha) – (1-cos^2beta)`
`=1/(1+cot^2alpha)-1+cos^2beta`
From (1) and (2),
`1/(1+1/4) -1+1 = 4/5`
So, option `d` is the correct option.