Question

If $\sin A=\frac{1}{3}$, then $\cos A.\csc A+\tan A.\sec A$ is equal to

• A. $\frac{16\sqrt{2}+3}{8}$
• B. $\frac{4\sqrt{2}+3}{8}$
• C. $\frac{\sqrt{3}+2}{4}$
• D. $\frac{\sqrt{3}-1}{8}$

Answer 1

The correct option is A $\frac{16\sqrt{2}+3}{8}$

$\sin A=\frac{1}{3}\therefore \cos A=\frac{1}{\sin A}=3$.

$\cos A=\sqrt{1-{\sin }^{2}A}=\sqrt{1-\frac{1}{9}}=\frac{\sqrt{8}}{3}=\frac{2\sqrt{2}}{3}$

$\tan A=\frac{1}{3}/\frac{\sqrt{8}}{3}=\frac{1}{2\sqrt{2}}.\sec A=\frac{1}{\cos A}=\frac{3}{2\sqrt{2}}$.

$\therefore \cos A\csc A+\tan A\sec A=\frac{2\sqrt{2}}{3}\times 3+\frac{1}{2\sqrt{2}}\times \frac{3}{2\sqrt{2}}$

$=2\sqrt{2}+\frac{3}{8}=\frac{16\sqrt{2}+3}{8}$ Ans