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 `3/5` (b) `3/5` (c) `2/(sqrt(5))` (d) `4/5`

Options

A. `(3)/(5)`
B. `(3)/(5)`
C. `(2)/sqrt(5)`
D. `(4)/(5)`

Answer

D

Explanation

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)`