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
\[\left( 1 \right){\text{ }}a = 1\]
\[\left( 2 \right){\text{ }}a = 2\]
\[\left( 3 \right){\text{ }}a = 3\]
\[\left( 4 \right){\text{ }}a = 4\]

Hint: 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 .