Question

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

• A. $\left(a\right)$ ${\tan }^{2}x=\frac{2}{3}$
• B. $\left(b\right)$ $\frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}=\frac{1}{125}$
• C. $\left(c\right)$ ${\tan }^{2}x=\frac{1}{3}$
• D. $\left(d\right)$ $\frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}=\frac{2}{125}$

Answer 1

Answer: (A), (B)

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

Ans. $\left(a\right)$, $\left(b\right)$
$3{\sin }^{4}x+2{(1-{\sin }^{2}x)}^2=6/5$
or $5{\sin }^{4}x-4{\sin }^{2}x+\frac{4}{5}=0$
or $25{\sin }^{4}x-20{\sin }^{2}x+4=0$
or ${(5{\sin }^{2}x-2)}^2=0\therefore {\sin }^{2}x=\frac{2}{5}$
$\therefore {\cos }^{2}x=1-{\sin }^{2}x=\frac{3}{5}\therefore {\tan }^{2}x=\frac{2}{3}$
Also $\frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}=\frac{1}{8}{\left(\frac{2}{5}\right)}^4+\frac{1}{27}{\left(\frac{3}{5}\right)}^4=\frac{1}{5^4}(2+3)=\frac{1}{5^3}=\frac{1}{125}$