Question

If $\frac{{\sin }^{4}x}{2}+\frac{{\cos }^{4}x}{3}=\frac{1}{5},$ then:

• A. ${\tan }^{2}x=\frac{2}{3}$
• B. $\frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}=\frac{1}{125}$
• C. ${\tan }^{2}x=\frac{1}{3}$
• D. $\frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}=\frac{2}{125}$

Answer 1

Answer: (A), (B)

The correct options are
A ${\tan }^{2}x=\frac{2}{3}$
B $\frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}=\frac{1}{125}$

If $\frac{{\sin }^{4}x}{2}+\frac{{\cos }^{4}x}{3}=\frac{1}{5},$ then

$\Rightarrow \frac{{\sin }^{4}x}{2}+\frac{{(1-{\sin }^{2}x)}^2}{3}=\frac{1}{5}$

$\Rightarrow \frac{{\sin }^{4}x}{2}+\frac{1+{\sin }^{4}x-2{\sin }^{2}x}{3}=\frac{1}{5}$

$\Rightarrow 25{\sin }^{4}x-20{\sin }^{2}x+4=0$

$\Rightarrow {(5{\sin }^{2}x-2)}^2=0\Rightarrow {\sin }^{2}x=\frac{2}{5}\Rightarrow {\cos }^{2}x=\frac{3}{5}$

$\Rightarrow {\tan }^{2}x=\frac{2}{3}$

and

$\frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}=\frac{16}{625\times 8}+\frac{81}{625\times 27}=\frac{1}{625}\times \frac{27\times 16+81\times 8}{8\times 27}=\frac{5}{625}=\frac{1}{125}.$