Question

If ${\sin }^{16}\alpha =\frac{1}{5}$, then the value of $\frac{1}{{\cos }^2\alpha }+\frac{1}{1+{\sin }^2\alpha }+\frac{2}{1+{\sin }^4\alpha }+\frac{4}{1+{\sin }^8\alpha }$ is

• A. $2$
• B. $4$
• C. $6$
• D. $8$
• E. $10$

Answer 1

The correct option is E $10$

Given, ${\sin }^{16}\alpha =\frac{1}{5}$ .... (i)
Now, $\frac{1}{{\cos }^2\alpha }+\frac{1}{1+{\sin }^2\alpha }+\frac{2}{1+{\sin }^4\alpha }+\frac{4}{1+{\sin }^8\alpha }$
$=\{\frac{1}{1-{\sin }^2\alpha }+\frac{1}{1+{\sin }^2\alpha }\}+\frac{2}{1+{\sin }^4\alpha }+\frac{4}{1+{\sin }^6\alpha }$
$=[\frac{2}{1-{\sin }^4\alpha }+\frac{2}{1+{\sin }^4\alpha }]+\frac{4}{1+{\sin }^6\alpha }$
$=2\cdot \left(\frac{2}{1-{\sin }^8\alpha }\right)+\frac{4}{1+{\sin }^8\alpha }$
$=4\left\{\frac{2}{1-{\sin }^{16}\alpha }\right\}=\frac{8}{1-\left(\frac{1}{5}\right)}=10$ ....[from Eq. (i)]