If \[\tan \alpha \] is equal to the integral solutions of the inequality \[4{x^2} - 16x + 15 < 0\] and \[\cos \beta \] is equal to the slope of the bisector of the first quadrant, then find the value of \[\sin \left( {\alpha + \beta } \right)\sin \left( {\alpha - \beta } \right)\].
A. \[\dfrac{3}{5}\]
B. \[\dfrac{2}{5}\]
C. \[\dfrac{2}{{\sqrt 5 }}\]
D. \[\dfrac{4}{5}\]

Hint: First we need to find the interval in which interval the value of the variable of inequality lied. Then we need to find the integral solution of inequality and after that we need to find the slope of the line with the given condition and simplify the given expression to get the required result.

Formula used
Trigonometric formula \[\sin (a + b)\sin (a - b) = {\cos ^2}b - {\cos ^2}a\]
\[\cos x = \dfrac{1}{{\sqrt {1 + {{\tan }^2}x} }}\]
General equation of slope intercept form of the straight line is \[y = mx + c\] , where \[m,\,c\] are constant.

Complete step by step solution:
Given inequality is \[4{x^2} - 16x + 15 < 0\]
Simplifying this inequality and we get
\[ \Rightarrow 4{x^2} - (10 + 6)x + 15 < 0\]
\[ \Rightarrow 4{x^2} - 10x - 6x + 15 < 0\]
\[ \Rightarrow 2x(2x - 5) - 3(2x - 5) < 0\]
\[ \Rightarrow (2x - 5)(2x - 3) < 0\]
\[ \Rightarrow \left( {x - \dfrac{5}{2}} \right)\left( {x - \dfrac{3}{2}} \right) < 0\]
Since the product of two expressions is less than zero, so one of them must be less than zero and other one positive.
Case I:
\[x - \dfrac{5}{2} < 0\] & \[x - \dfrac{3}{2} > 0\]
\[x < \dfrac{5}{2}\] & \[x > \dfrac{3}{2}\]
So, \[x \in \left( {\dfrac{3}{2},\dfrac{5}{2}} \right)\]
Case II:
\[x - \dfrac{5}{2} > 0\] & \[x - \dfrac{3}{2} < 0\]
\[x > \dfrac{5}{2}\] & \[x < \dfrac{3}{2}\]
The second case is impossible , so considering the first case.
\[x \in \left( {\dfrac{3}{2},\dfrac{5}{2}} \right)\]
Given that \[\tan \alpha \] is equal to the integral solutions of the inequality \[4{x^2} - 16x + 15 < 0\]
Therefore, \[\tan \alpha = 2\]
Now for the second condition,
The bisector of the first quadrant is the straight line \[y = x\]………….(1)
The general equation of slope \[y = mx + c\] …………..(2)
Comparing (1) and (2) and we get
\[m = 1\] and \[c = 0\]
Given that \[\cos \beta \] is equal to the slope of the bisector of the first quadrant, therefore
\[\cos \beta = 1\]
Now \[\sin \left( {\alpha + \beta } \right)\sin \left( {\alpha - \beta } \right)\] \[ = {\cos ^2}\beta - {\cos ^2}\alpha \]
Now we use the property of cosine \[\cos x = \dfrac{1}{{\sqrt {1 + {{\tan }^2}x} }}\] and we get
\[ = {\cos ^2}\beta - {\left( {\dfrac{1}{{\sqrt {1 + {{\tan }^2}\alpha } }}} \right)^2}\]
Now substitute the values of tangent and cosecant and we get
\[ = {\left( 1 \right)^2} - {\left( {\dfrac{1}{{\sqrt {1 + {{\left( 2 \right)}^2}} }}} \right)^2}\]
Simplifying and we get
\[ = 1 - {\left( {\dfrac{1}{{\sqrt {1 + 4} }}} \right)^2}\]
\[ = 1 - {\left( {\dfrac{1}{{\sqrt 5 }}} \right)^2}\]
\[ = 1 - \dfrac{1}{5}\]
\[ = \dfrac{{5 - 1}}{5}\]
\[ = \dfrac{4}{5}\]
Hence, the correct option is option D.

Note: We need to take care while we use the properties of cosecant in the given expression to find the required result. If we use different properties instead of the conversion of cosecant to tangent then it will be tough to get the required result.