If a,b,c are positive constants such that `agtb` then the maximum value of `r` , given by `c^4/r^2=a^2/(sin^2theta)+b^2/(cos^2theta)` , must be

Options

A. `c^2/(a-b)`
B. `c^2/(a+b)`
C. `c^2/(sin^2theta)`
D. `c^2/(sqrt(ab))`

Answer

To solve the problem, we need to find the maximum value of \( r \) given the equation:

\[ \frac{c^4}{r^2} = \frac{a^2}{\sin^2 \theta} + \frac{b^2}{\cos^2 \theta} \]

### Step-by-step Solution:

1. **Rearranging the Equation:**
Start by rearranging the equation to express \( r^2 \):

\[ r^2 = \frac{c^4}{\frac{a^2}{\sin^2 \theta} + \frac{b^2}{\cos^2 \theta}} \]

2. **Using Trigonometric Identities:**
We can rewrite the terms involving \( \sin^2 \theta \) and \( \cos^2 \theta \):

\[ \frac{a^2}{\sin^2 \theta} + \frac{b^2}{\cos^2 \theta} = a^2 \csc^2 \theta + b^2 \sec^2 \theta \]

3. **Applying the Cauchy-Schwarz Inequality:**
To find the minimum value of the denominator, we can apply the Cauchy-Schwarz inequality:

\[ (a^2 \sin^2 \theta + b^2 \cos^2 \theta) \left(\frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta}\right) \geq (a + b)^2 \]

This implies:

\[ \frac{a^2}{\sin^2 \theta} + \frac{b^2}{\cos^2 \theta} \geq \frac{(a + b)^2}{\sin^2 \theta + \cos^2 \theta} = (a + b)^2 \]

4. **Substituting Back:**
Substitute back into the expression for \( r^2 \):

\[ r^2 \leq \frac{c^4}{(a + b)^2} \]

5. **Taking the Square Root:**
Taking the square root gives us the maximum value of \( r \):

\[ r \leq \frac{c^2}{a + b} \]

6. **Conclusion:**
Therefore, the maximum value of \( r \) is:

\[ r_{\text{max}} = \frac{c^2}{a + b} \]

### Final Answer:
The maximum value of \( r \) is \( \frac{c^2}{a + b} \).