What is the minimum value of `4cosec^(2)alpha+9sin^(2)alpha` ?

Options

A. `10`
B. `11`
C. `12`
D. `14`

Answer

To find the minimum value of the expression \(4 \csc^2 \alpha + 9 \sin^2 \alpha\), we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the expression in a more manageable form. We can express \(4 \csc^2 \alpha\) as \(4 \cdot \frac{1}{\sin^2 \alpha}\):
\[ 4 \csc^2 \alpha + 9 \sin^2 \alpha = \frac{4}{\sin^2 \alpha} + 9 \sin^2 \alpha \]

### Step 2: Introduce a substitution
Let \(x = \sin^2 \alpha\). Then, we can rewrite the expression as:
\[ f(x) = \frac{4}{x} + 9x \]
where \(0 < x \leq 1\) since \(\sin^2 \alpha\) ranges from 0 to 1.

### Step 3: Find the derivative
To find the minimum value, we need to differentiate \(f(x)\) and set the derivative to zero:
\[ f'(x) = -\frac{4}{x^2} + 9 \]
Setting \(f'(x) = 0\):
\[ -\frac{4}{x^2} + 9 = 0 \]
\[ \frac{4}{x^2} = 9 \]
\[ x^2 = \frac{4}{9} \]
\[ x = \frac{2}{3} \]

### Step 4: Evaluate the second derivative
To confirm that this point is a minimum, we can check the second derivative:
\[ f''(x) = \frac{8}{x^3} \]
Since \(f''(x) > 0\) for \(x > 0\), this indicates that \(f(x)\) is concave up and thus \(x = \frac{2}{3}\) is indeed a minimum.

### Step 5: Calculate the minimum value
Now we substitute \(x = \frac{2}{3}\) back into the original function:
\[ f\left(\frac{2}{3}\right) = \frac{4}{\frac{2}{3}} + 9 \cdot \frac{2}{3} \]
\[ = 4 \cdot \frac{3}{2} + 6 = 6 + 6 = 12 \]

### Conclusion
The minimum value of \(4 \csc^2 \alpha + 9 \sin^2 \alpha\) is \(12\).

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