Question

If $15{\sin }^4\left(\theta \right)+10{\cos }^4\left(\theta \right)=6,$ for some $\theta \in R$ then the value of $27{\sec }^6\left(\theta \right)+8{\mathrm{cosec}}^6\left(\theta \right)$ is equal to:

• A. $250$
• B. $500$
• C. $400$
• D. $350$

Answer 1

The correct option is A

$250$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $15{\sin }^4\left(\theta \right)+10{\cos }^4\left(\theta \right)=6,$

$\Rightarrow$ $15{\sin }^4\left(\theta \right)+10{\left(1-{\sin }^2\left(\theta \right)\right)}^2=6,$

$\Rightarrow$$15{\sin }^4\left(\theta \right)+10\left({\left(1\right)}^2+{\sin }^4\left(\theta \right)-2{\sin }^2\left(\theta \right)\right)=6,$

$\Rightarrow$ $25{\sin }^4\left(\theta \right)-20{\sin }^2\left(\theta \right)+4=0,$

$\Rightarrow$ ${\left(5{\sin }^2\left(\theta \right)\right)}^2-2\times 5\times 2\times {\sin }^2\left(\theta \right)+{\left(2\right)}^2=0,$

$\Rightarrow$ ${\left(5{\sin }^2\left(\theta \right)-2\right)}^2=0$

$\Rightarrow$ ${\sin }^2\left(\theta \right)=\frac{2}{5}$

$\Rightarrow$ ${\cos }^2\left(\theta \right)=1-\frac{2}{5}$

$\Rightarrow$ ${\cos }^2\left(\theta \right)=\frac{3}{5}$

Step-2: Finding the value of $27{\sec }^6\left(\theta \right)+8{\mathrm{cosec}}^6\left(\theta \right)$.

$\Rightarrow$$27{\sec }^6\left(\theta \right)+8{\mathrm{cosec}}^6\left(\theta \right)=27\times {\left(\frac{5}{3}\right)}^3+8\times {\left(\frac{5}{2}\right)}^3$

$\Rightarrow$$27{\sec }^6\left(\theta \right)+8{\mathrm{cosec}}^6\left(\theta \right)=27\times \frac{125}{27}+8\times \frac{125}{8}$

$\Rightarrow$$27{\sec }^6\left(\theta \right)+8{\mathrm{cosec}}^6\left(\theta \right)=250$

Hence, correct answer is option (A).