The least values of `cosec^(2) x + 25 sec^(2)x` is

Options

A. 0
B. 26
C. 28
D. 36

Answer

To find the least value of the expression \( \csc^2 x + 25 \sec^2 x \), we will follow these steps:

### Step 1: Rewrite the expression
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 the expression:
\[ \csc^2 x + 25 \sec^2 x = (1 + \cot^2 x) + 25(1 + \tan^2 x) \]
This simplifies to:
\[ = 1 + \cot^2 x + 25 + 25 \tan^2 x = 26 + \cot^2 x + 25 \tan^2 x \]

### Step 2: Let \( a = \cot^2 x \) and \( b = \tan^2 x \)
From the identity \( \cot^2 x = \frac{1}{\tan^2 x} \), we can denote:
\[ a = \cot^2 x \quad \text{and} \quad b = \tan^2 x \]
Thus, we have:
\[ \cot^2 x + 25 \tan^2 x = a + 25b \]

### Step 3: Use the identity \( ab = 1 \)
Since \( a = \frac{1}{b} \), we can express \( a \) in terms of \( b \):
\[ a = \frac{1}{b} \]
Substituting this into our expression:
\[ \frac{1}{b} + 25b \]

### Step 4: Find the minimum value using AM-GM inequality
To find the minimum value of \( \frac{1}{b} + 25b \), we apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{\frac{1}{b} + 25b}{2} \geq \sqrt{\frac{1}{b} \cdot 25b} \]
This simplifies to:
\[ \frac{\frac{1}{b} + 25b}{2} \geq 5 \]
Thus:
\[ \frac{1}{b} + 25b \geq 10 \]

### Step 5: Add the constant term
Now, we add the constant term we factored out earlier:
\[ 26 + \left(\frac{1}{b} + 25b\right) \geq 26 + 10 = 36 \]

### Step 6: Conclusion
The least value of \( \csc^2 x + 25 \sec^2 x \) is:
\[ \boxed{36} \]

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