Question

The minimum value of ${\sec ^2}x + \cos e{c^2}x$ equals the maximum value of $a{\sin ^2}x + b{\cos ^2}x$ where $a > b > 0$ . The value of $a$ is
$\left( 1 \right){\text{ }}a = 1$
$\left( 2 \right){\text{ }}a = 2$
$\left( 3 \right){\text{ }}a = 3$
$\left( 4 \right){\text{ }}a = 4$

Answer 1

In this we will use the tricks to find the minimum and maximum values of trigonometric identities like for $a{\text{ta}}{{\text{n}}^2}x + b{\cot ^2}x$ the minimum value is $2\sqrt {ab}$ . But first we have to deduce the equation ${\sec ^2}x + \cos e{c^2}x$ and then we can apply the trick. Then find the maximum value of $a{\sin ^2}x + b{\cos ^2}x$ and $\sin x$ is maximum at $x = \dfrac{\pi }{2}$ . Then use the condition given in the question to find out the value of a.

Complete step by step answer:
Our step is to find the minimum value of the equation ${\sec ^2}x + \cos e{c^2}x$ . Because ${\sec ^2}x{\text{ }} = {\text{ 1 + ta}}{{\text{n}}^2}x$ and $\cos e{c^2}x{\text{ }} = {\text{ }}1 + {\cot ^2}x$ . Therefore,
${\sec ^2}x + \cos e{c^2}x = {\text{ 1 + ta}}{{\text{n}}^2}x + 1 + {\cot ^2}x$
$= {\text{ 2 + ta}}{{\text{n}}^2}x + {\cot ^2}x$
Now because the minimum value of $a{\text{ta}}{{\text{n}}^2}x + b{\cot ^2}x = 2\sqrt {ab}$ and here the values of $a$ and $b$ is 1. Therefore,
$= {\text{ 2 + }}2\sqrt {1 \times 1}$
Further simplifying we get,
$= {\text{ 2 + }}2$
$= {\text{ 4}}$
From this we have the minimum value of the equation ${\sec ^2}x + \cos e{c^2}x$ as ${\text{4}}$ .
Next our second step is to find out the maximum value of $a{\sin ^2}x + b{\cos ^2}x$ . We know that ${\sin ^2}x + {\cos ^2}x = 1$ , so ${\cos ^2}x = 1 - {\sin ^2}x$ . Therefore,
$a{\sin ^2}x + b{\cos ^2}x = {\text{ }}a{\sin ^2}x + b\left( {1 - {{\sin }^2}x} \right)$
$= {\text{ }}a{\sin ^2}x + b - b{\sin ^2}x$
By taking out ${\sin ^2}x$ common we get
$= {\text{ }}\left( {a - b} \right){\sin ^2}x + b$
It is given that $a$ is greater than $b$ . Therefore there exists a maximum value at $\sin x = 1$ that is if the value of $x$ is $\dfrac{\pi }{2}$ . So,
$= {\text{ }}\left( {a - b} \right){\left( 1 \right)^2} + b$
$= {\text{ }}a - b + b$
The terms $b$ will cancels each other
$= {\text{ }}a$
Thus, the maximum value of $a{\sin ^2}x + b{\cos ^2}x$ is $a$
It is also given that minimum value of ${\sec ^2}x + \cos e{c^2}x$ $=$ maximum value of $a{\sin ^2}x + b{\cos ^2}x$ .
$\therefore$ $4{\text{ }} = {\text{ }}a$
$\Rightarrow a{\text{ }} = {\text{ 4}}$

So, the correct answer is “Option 4”.

Note:
In general for $a{\sec ^2}x + b\cos e{c^2}x$ the minimum value is $a + b + 2\sqrt {ab}$ . Remember that first we have to deduce the equation up to the point we can. In $a{\sin ^2}x + b{\cos ^2}x$ , if $a > b$ then the maximum value is a and the minimum value is b but if $b > a$ then the maximum value is b and the minimum value is a. Also note that the value of sine and cosine remains between $1$ and $- 1$ . So clearly their maximum value is $1$ . In general, the maximum value of $A\sin x$ is $A$ . And as we know that sec and cosec are the reciprocal of cosine and sine, so their values also vary with sine and cosine but inversely .