Find `max^(m)&min^(m)` value of `23sec^(2)theta+20tan^(2)theta`

Answer

To find the maximum and minimum values of the expression \(23 \sec^2 \theta + 20 \tan^2 \theta\), we can follow these steps:

### Step 1: Use the identity for secant and tangent
We know that:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
From this, we can express \(\sec^2 \theta\) in terms of \(\tan^2 \theta\):
\[ \sec^2 \theta = 1 + \tan^2 \theta \]

### Step 2: Substitute \(\sec^2 \theta\) in the expression
Now, substitute \(\sec^2 \theta\) into the original expression:
\[ 23 \sec^2 \theta + 20 \tan^2 \theta = 23(1 + \tan^2 \theta) + 20 \tan^2 \theta \]
This simplifies to:
\[ 23 + 23 \tan^2 \theta + 20 \tan^2 \theta = 23 + 43 \tan^2 \theta \]

### Step 3: Analyze the expression
Now we have:
\[ f(x) = 23 + 43 x \]
where \(x = \tan^2 \theta\). Since \(\tan^2 \theta\) can take values from \(0\) to \(\infty\) (as \(\theta\) varies), we can analyze the behavior of \(f(x)\).

### Step 4: Determine the minimum value
The minimum value occurs when \(\tan^2 \theta = 0\):
\[ f(0) = 23 + 43 \cdot 0 = 23 \]

### Step 5: Determine the maximum value
As \(\tan^2 \theta\) approaches infinity, \(f(x)\) also approaches infinity:
\[ \lim_{x \to \infty} f(x) = 23 + 43 \cdot \infty = \infty \]

### Conclusion
Thus, the minimum value of the expression \(23 \sec^2 \theta + 20 \tan^2 \theta\) is \(23\) and the maximum value is \(\infty\).

### Final Answer
- Minimum value: \(23\)
- Maximum value: \(\infty\)