If tan alpha equals the integral solution of | vedclass

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(D) Given the inequality $4x^2 - 16x + 15 < 0$.Solving for $x$: $(2x - 3)(2x - 5) < 0$,which gives $\frac{3}{2} < x < \frac{5}{2}$.The only integral s

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$\frac{4}{5}$

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(D) Given the inequality $4x^2 - 16x + 15 Solving for $x$: $(2x - 3)(2x - 5) The only integral solution in this interval is $x = 2$. Thus,$	an \alpha = 2$.The bisector of the first quadrant is the line $y = x$,which has a slope of $1$. Thus,$\cos \beta = 1$.Since $\cos \beta = 1$,we have $\beta = 0$,so $\sin \beta = 0$.Using the identity $\sin(\alpha + \beta)\sin(\alpha - \beta) = \sin^2 \alpha - \sin^2 \beta$:$\sin^2 \alpha - \sin^2 \beta = \sin^2 \alpha - 0 = \sin^2 \alpha$.Since $	an \alpha = 2$,we have $\sin^2 \alpha = \frac{	an^2 \alpha}{1 + 	an^2 \alpha} = \frac{2^2}{1 + 2^2} = \frac{4}{5}$.

If $\tan \alpha$ equals the integral solution of the inequality $4x^2 - 16x + 15 < 0$ and $\cos \beta$ equals the slope of the bisector of the first quadrant,then $\sin(\alpha + \beta)\sin(\alpha - \beta)$ is equal to

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