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 $\frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}=\frac{1}{125}$
B ${\tan }^{2}x=\frac{2}{3}$

Given that $\frac{{\sin }^{4}x}{2}+\frac{{\cos }^{4}x}{3}=\frac{1}{5}$
$\Rightarrow 3{\sin }^{4}x+2{\cos }^{4}x=\frac{6}{5}$
$\Rightarrow {\sin }^{4}x+2[{\sin }^{4}x+{\cos }^{4}x]=\frac{6}{5}$

$\Rightarrow {\sin }^{4}x+2[{\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x-2{\sin }^{2}x{\cos }^{2}x]=\frac{6}{5}$

$\Rightarrow {\sin }^{4}x+2\left[\right({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x]=\frac{6}{5}$
$\Rightarrow {\sin }^{4}x+2[1-2{\sin }^{2}x{\cos }^{2}x]=\frac{6}{5}$
$\Rightarrow {\sin }^{4}x+2-4{\sin }^{2}x(1-{\sin }^{2}x)=\frac{6}{5}$
$\Rightarrow 25{\sin }^{4}x-20{\sin }^{2}x+10-6=0$
$\Rightarrow (5{\sin }^{2}x-2{)}^2=0$
$\Rightarrow {\cos }^{2}x=\frac{3}{5}$ and ${\tan }^{2}x=\frac{2}{3}$
Also $\frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}=\frac{2}{625}+\frac{3}{625}=\frac{5}{625}=\frac{1}{125}$