Find the range of `f(x)=cosec^2x+25sec^2x`.

Explanation

To find the range of the function \( f(x) = \csc^2 x + 25 \sec^2 x \), we can follow these steps:

### Step 1: Rewrite the function using trigonometric identities
We know that:
\[ \csc^2 x = 1 + \cot^2 x \quad \text{and} \quad \sec^2 x = 1 + \tan^2 x \]
Thus, we can rewrite \( f(x) \) as:
\[ f(x) = \csc^2 x + 25 \sec^2 x = (1 + \cot^2 x) + 25(1 + \tan^2 x) \]
This simplifies to:
\[ f(x) = 1 + \cot^2 x + 25 + 25 \tan^2 x = 26 + \cot^2 x + 25 \tan^2 x \]

### Step 2: Substitute \( y = \cot x \) and \( z = \tan x \)
Using the identity \( \cot^2 x = \frac{1}{\tan^2 x} \), we can let \( y = \cot x \) and \( z = \tan x \). Note that \( y = \frac{1}{z} \). Therefore, we can express \( f(x) \) in terms of \( z \):
\[ f(z) = 26 + y^2 + 25z^2 = 26 + \frac{1}{z^2} + 25z^2 \]

### Step 3: Find the minimum value of the function
To find the minimum value of \( f(z) \), we can differentiate it with respect to \( z \) and set the derivative to zero. However, it is more straightforward to analyze the expression:
\[ f(z) = 26 + \frac{1}{z^2} + 25z^2 \]
Let \( u = z^2 \), then:
\[ f(u) = 26 + \frac{1}{u} + 25u \]
Now we can find the critical points by taking the derivative:
\[ f'(u) = -\frac{1}{u^2} + 25 \]
Setting \( f'(u) = 0 \):
\[ -\frac{1}{u^2} + 25 = 0 \implies \frac{1}{u^2} = 25 \implies u^2 = \frac{1}{25} \implies u = \frac{1}{5} \]

### Step 4: Calculate the minimum value
Substituting \( u = \frac{1}{5} \) back into \( f(u) \):
\[ f\left(\frac{1}{5}\right) = 26 + 25\left(\frac{1}{5}\right) + 5 = 26 + 5 + 25 = 36 \]
Thus, the minimum value of \( f(x) \) is 36.

### Step 5: Determine the range of the function
Since \( \frac{1}{z^2} \) approaches infinity as \( z \) approaches 0, and \( 25z^2 \) approaches infinity as \( z \) approaches infinity, the function \( f(z) \) can take values from its minimum of 36 to infinity.

### Conclusion
The range of the function \( f(x) = \csc^2 x + 25 \sec^2 x \) is:
\[ \text{Range of } f(x) = [36, \infty) \]