Question

If $\frac{{\sin }^{4}x}{3}+\frac{{\cos }^{4}x}{4}=\frac{1}{7},$ then the value of $\frac{{\sec }^{2}x}{3}+\frac{{\text{cosec}}^{2}x}{4}$ is

• A. $\frac{12}{7}$
• B. $\frac{7}{6}$
• C. $\frac{16}{7}$
• D. $\frac{13}{6}$

Answer 1

The correct option is B $\frac{7}{6}$

$\frac{{\sin }^{4}x}{3}+\frac{{\cos }^{4}x}{4}=\frac{1}{7}\Rightarrow \frac{7{\sin }^{4}x}{3}+\frac{7{\cos }^{4}x}{4}=1\Rightarrow {\sin }^{4}x+\frac{4}{3}{\sin }^{4}x+{\cos }^{4}x+\frac{3}{4}{\cos }^{4}x=1\Rightarrow 1-2{\sin }^{2}x{\cos }^{2}x+\frac{4}{3}{\sin }^{4}x+\frac{3}{4}{\cos }^{4}x=1(\because {\sin }^{4}x+{\cos }^{4}x=1-2{\sin }^{2}x{\cos }^{2}x)$
$\Rightarrow {(\frac{2}{\sqrt{3}}{\sin }^{2}x-\frac{\sqrt{3}}{2}{\cos }^{2}x)}^2=0$
$\Rightarrow {\tan }^{2}x=\frac{3}{4}$

$\therefore \frac{{\sec }^{2}x}{3}+\frac{{\text{cosec}}^{2}x}{4}$
$=\frac{1+{\tan }^{2}x}{3}+\frac{1+{\cot }^{2}x}{4}=\frac{7}{12}+\frac{7}{12}=\frac{7}{6}$