question_answer

If sin \[A=\frac{1}{3},\] then \[\cos A\text{cosec}\,A+\tan A\sec A\] is equal to

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, \[\sin A=\frac{1}{3}\] \[A{{B}^{2}}=A{{C}^{2}}-B{{C}^{2}}\]         \[={{3}^{2}}-1=9-1\] \[A{{B}^{2}}=8\] \[AB=2\sqrt{2}\]         \[\therefore \]      \[\cos A=\frac{2\sqrt{2}}{3},\] \[\sec A=\frac{3}{2\sqrt{2}}\]             \[\tan A=\frac{1}{2\sqrt{2}}\]             \[\text{cosec}\,\text{A=}\frac{1}{\sin A}=3\] \[\therefore \]      \[\cos A\cdot cosec\,A+tan\,A\cdot sec\,A\]             \[=\frac{2\sqrt{2}}{3}\cdot \frac{3}{1}+\frac{1}{2\sqrt{2}}\cdot \frac{3}{2\sqrt{2}}\]             \[=2\sqrt{2}+\frac{3}{8}=\frac{16\sqrt{2}+3}{8}\]