Question

The minimum value of ${\sec }^2\left(x\right)+{\mathrm{cosec}}^2\left(x\right)$ equals maximum value of $a{\sin }^2\left(x\right)+b{\cos }^2\left(x\right)$, where $a>b>0$, the value of $a$ is

• A. $a=1$
• B. $a=2$
• C. $a=3$
• D. $a=4$

Answer 1

The correct option is D

$a=4$

Explanation for the correct option:

Step 1: Find the minimum value of first equation

Given functions are, ${\sec }^2\left(x\right)+{\mathrm{cosec}}^2\left(x\right)$ and $a{\sin }^2\left(x\right)+b{\cos }^2\left(x\right)$

We have,
${\sec }^2\left(x\right)+{\mathrm{cosec}}^2\left(x\right)=2+{\tan }^2\left(x\right)+{\cot }^2\left(x\right)\left[\because {\sec }^2\left(x\right)=1+{\tan }^2\left(x\right),{\mathrm{cosec}}^2\left(x\right)=1+{\cot }^2\left(x\right)\right]$

We know that,
$a{\tan }^2\left(x\right)+b{\cot }^2\left(x\right)\ge 2\sqrt{\mathrm{ab}}$

Comparing with the given equation,
$a=1$ and $b=1$

Thus, the minimum value of $2+{\tan }^2\left(x\right)+{\cot }^2\left(x\right)$ is,
$2+2\sqrt{1\times 1}=4$

Step 2: Find the maximum value of second equation

We know that,
${\sin }^2\left(x\right)+{\cos }^2\left(x\right)=1$

So,
$\begin{array}{rcl}a{\sin }^2\left(x\right)+b{\cos }^2\left(x\right)& =& a{\sin }^2\left(x\right)+b\left[1-{\sin }^2\left(x\right)\right]\\ & =& a{\sin }^2\left(x\right)+b-b{\sin }^2\left(x\right)\\ & =& \left(a-b\right){\sin }^{2}x+b\end{array}$

We see that the function is maximum when ${\sin }^2\left(x\right)=1$ which happens at $x=\frac{\mathrm{\pi }}{2}$, since $a>b$
i.e., $\left(a-b\right)+b=a$

So the maximum value of $a{\sin }^2\left(x\right)+b{\cos }^2\left(x\right)$ is $a$

Step 3: Set equal the maximum and minimum value of the equations

From given,
$\begin{array}{cc}\therefore & \min \left[{\sec }^2\left(x\right)+{\mathrm{cosec}}^2\left(x\right)\right]=\max \left[a{\sin }^2\left(\mathrm{x}\right)+b{\cos }^2\left(x\right)\right]\\ \Rightarrow & 4=a\\ \therefore & a=4\end{array}$

Therefore, the value of $a$ is $4$

Hence, option A is correct.