If `tanalpha` is equal to the integral solution of the inequality `4x^2-16 x+15<0` and `cosbeta` is equal to the slope of the bisector of the first quadrant, then `sin(alpha+beta)sin(alpha-beta)` is equal to (a)`3/5` (b) `3/5` (c) `2/(sqrt(5))` (d) `4/5`

Explanation

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,

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