The least value of `18 sin^(2)theta+2 cosec^(2)theta-3` is (a) -15 (b) -12 (c) 0 (d) 9

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:
\[ f(\theta) = 18 \sin^2 \theta + 2 \csc^2 \theta - 3 \]
Recall that \( \csc^2 \theta = \frac{1}{\sin^2 \theta} \). Thus, we can rewrite the expression as:
\[ f(\theta) = 18 \sin^2 \theta + 2 \cdot \frac{1}{\sin^2 \theta} - 3 \]

### Step 2: Substitute \( x = \sin^2 \theta \)
Let \( x = \sin^2 \theta \). Then, \( \csc^2 \theta = \frac{1}{x} \) and the expression becomes:
\[ f(x) = 18x + \frac{2}{x} - 3 \]
where \( 0 < x \leq 1 \).

### Step 3: Find the derivative
To find the minimum value, we first find the derivative of \( f(x) \):
\[ f'(x) = 18 - \frac{2}{x^2} \]
Setting the derivative equal to zero to find critical points:
\[ 18 - \frac{2}{x^2} = 0 \]
\[ \frac{2}{x^2} = 18 \]
\[ x^2 = \frac{2}{18} = \frac{1}{9} \]
\[ x = \frac{1}{3} \quad (\text{since } x > 0) \]

### Step 4: Evaluate the second derivative
To confirm that this point is a minimum, we compute the second derivative:
\[ f''(x) = \frac{4}{x^3} \]
Since \( f''(x) > 0 \) for \( x > 0 \), the function is concave up, confirming a local minimum at \( x = \frac{1}{3} \).

### Step 5: Calculate the minimum value
Now, we substitute \( x = \frac{1}{3} \) back into the original expression:
\[ f\left(\frac{1}{3}\right) = 18 \left(\frac{1}{3}\right) + \frac{2}{\frac{1}{3}} - 3 \]
\[ = 6 + 6 - 3 \]
\[ = 9 \]

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

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