The least value of `18 sin^(2)theta+2 cosec^(2)theta-3` is

Options

A. `-15`
B. `-12`
C. 0
D. 9

Explanation

To find the least value of the expression \( 18 \sin^2 \theta + 2 \csc^2 \theta - 3 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ f(\theta) = 18 \sin^2 \theta + 2 \csc^2 \theta - 3 \]
Since \(\csc^2 \theta = \frac{1}{\sin^2 \theta}\), we can rewrite the expression as:
\[ f(\theta) = 18 \sin^2 \theta + \frac{2}{\sin^2 \theta} - 3 \]

### Step 2: Let \( x = \sin^2 \theta \)
Let \( x = \sin^2 \theta \). Then, \( 0 < x \leq 1 \) and we can rewrite the function as:
\[ f(x) = 18x + \frac{2}{x} - 3 \]

### Step 3: Find the derivative
Next, we find the derivative of \( f(x) \):
\[ f'(x) = 18 - \frac{2}{x^2} \]

### Step 4: Set the derivative to zero
To find the critical points, we set the derivative to zero:
\[ 18 - \frac{2}{x^2} = 0 \]
Solving for \( x \):
\[ \frac{2}{x^2} = 18 \implies x^2 = \frac{2}{18} \implies x^2 = \frac{1}{9} \implies x = \frac{1}{3} \]

### Step 5: Check the second derivative
Now, we check the second derivative to confirm if this point is a minimum:
\[ f''(x) = \frac{4}{x^3} \]
Since \( f''(x) > 0 \) for \( x > 0 \), this indicates that \( x = \frac{1}{3} \) is indeed a minimum.

### Step 6: Calculate the minimum value
Now, we substitute \( x = \frac{1}{3} \) back into the function to find the minimum value:
\[ f\left(\frac{1}{3}\right) = 18\left(\frac{1}{3}\right) + \frac{2}{\frac{1}{3}} - 3 \]
Calculating this gives:
\[ = 6 + 6 - 3 = 9 \]

### Conclusion
Thus, the least value of the expression \( 18 \sin^2 \theta + 2 \csc^2 \theta - 3 \) is:
\[ \boxed{9} \]