A) \[\frac{3}{5}\]
B) \[-\frac{3}{5}\]
C) \[\frac{2}{\sqrt{5}}\]
D) \[\frac{4}{5}\]
Correct Answer: D
Solution :
We have \[4{{x}^{2}}-16x+15<0\,\,\Rightarrow \,\frac{3}{2}<x<\frac{5}{2}\] \[\therefore \] Integral solution of \[4{{x}^{2}}-16x+15<0\] is x = 2. Thus \[\tan \alpha =2.\] It is given that \[\cos \beta =\tan {{45}^{o}}=1\] \[\therefore \,\,\sin \,(\alpha +\beta )\,\sin \,(\alpha -\beta )={{\sin }^{2}}\alpha -{{\sin }^{2}}\beta \] \[=\frac{1}{1+{{\cot }^{2}}\alpha }-(1-{{\cos }^{2}}\beta )=\frac{1}{1+\frac{1}{4}}-0=\frac{4}{5}\].
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