A) \[\frac{16\sqrt{2}+3}{8}\]
B) \[\frac{4\sqrt{2}+3}{8}\]
C) \[\frac{\sqrt{3}+2}{8}\]
D) \[\frac{\sqrt{3}-1}{8}\]
Correct Answer: A
Solution :
Here, \[\sin A=\frac{1}{3}\] \[A{{B}^{2}}=A{{C}^{2}}-B{{C}^{2}}\] \[={{3}^{2}}-1=9-1\] \[A{{B}^{2}}=8\] \[AB=2\sqrt{2}\] \[\therefore \] \[\cos A=\frac{2\sqrt{2}}{3},\] \[\sec A=\frac{3}{2\sqrt{2}}\] \[\tan A=\frac{1}{2\sqrt{2}}\] \[\text{cosec}\,\text{A=}\frac{1}{\sin A}=3\] \[\therefore \] \[\cos A\cdot cosec\,A+tan\,A\cdot sec\,A\] \[=\frac{2\sqrt{2}}{3}\cdot \frac{3}{1}+\frac{1}{2\sqrt{2}}\cdot \frac{3}{2\sqrt{2}}\] \[=2\sqrt{2}+\frac{3}{8}=\frac{16\sqrt{2}+3}{8}\]You need to login to perform this action.
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