trigonometry

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



1) $\frac{3}{5}$
2) $-\frac{3}{5}$
3) $\frac{2}{\sqrt{5}}$
4) $\frac{4}{5}$
Solution: Explanation: No Solution
Trigonometric ratios of sum and difference of two and three

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