The least values of `( sin ^(2) theta+ cosec ^(2) theta)` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

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

### Step 1: Rewrite the expression
We know that \( \csc^2 \theta = \frac{1}{\sin^2 \theta} \). Therefore, we can rewrite the expression as:
\[ \sin^2 \theta + \csc^2 \theta = \sin^2 \theta + \frac{1}{\sin^2 \theta} \]

### Step 2: Let \( x = \sin^2 \theta \)
Now, we can let \( x = \sin^2 \theta \). The expression becomes:
\[ f(x) = x + \frac{1}{x} \]
where \( x \) is in the interval \( (0, 1] \) since \( \sin^2 \theta \) ranges from 0 to 1.

### Step 3: Find the derivative
To find the minimum value, we can differentiate \( f(x) \):
\[ f'(x) = 1 - \frac{1}{x^2} \]

### Step 4: Set the derivative to zero
Setting the derivative equal to zero to find critical points:
\[ 1 - \frac{1}{x^2} = 0 \implies \frac{1}{x^2} = 1 \implies x^2 = 1 \implies x = 1 \]

### Step 5: Evaluate the second derivative
Now we check the second derivative to confirm if it is a minimum:
\[ f''(x) = \frac{2}{x^3} \]
Since \( f''(x) > 0 \) for \( x > 0 \), this indicates that \( f(x) \) is concave up at \( x = 1 \).

### Step 6: Calculate the minimum value
Now, we can calculate the value of \( f(x) \) at \( x = 1 \):
\[ f(1) = 1 + \frac{1}{1} = 2 \]

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