`sec^(2)theta==(4ab)/((a+b)^(2)),` where a, b `inR` is true if and olny if

Options

A. `a+b ne0`
B. `a=b, a ne0`
C. `a=b`
D. `a ne 0, bne0`

Explanation

To solve the equation \( \sec^2 \theta = \frac{4ab}{(a+b)^2} \) where \( a, b \in \mathbb{R} \), we will follow these steps:

### Step 1: Understand the properties of \( \sec^2 \theta \)
We know that \( \sec^2 \theta \) is always greater than or equal to 1 for all real values of \( \theta \). Therefore, we can write:
\[ \sec^2 \theta \geq 1 \]

### Step 2: Set up the inequality
Given that \( \sec^2 \theta = \frac{4ab}{(a+b)^2} \), we can substitute this into our inequality:
\[ \frac{4ab}{(a+b)^2} \geq 1 \]

### Step 3: Rearrange the inequality
To eliminate the fraction, we multiply both sides by \( (a+b)^2 \) (noting that \( (a+b)^2 > 0 \) since \( a, b \) are real numbers):
\[ 4ab \geq (a+b)^2 \]

### Step 4: Expand the right-hand side
Now, we expand \( (a+b)^2 \):
\[ 4ab \geq a^2 + 2ab + b^2 \]

### Step 5: Rearrange the terms
Rearranging gives us:
\[ 4ab - 2ab - a^2 - b^2 \geq 0 \]
which simplifies to:
\[ 2ab - a^2 - b^2 \geq 0 \]

### Step 6: Factor the inequality
We can rewrite this as:
\[ -a^2 + 2ab - b^2 \geq 0 \]
This can be factored as:
\[ -(a^2 - 2ab + b^2) \geq 0 \]
or equivalently:
\[ -(a-b)^2 \geq 0 \]

### Step 7: Analyze the inequality
Since \( -(a-b)^2 \) is less than or equal to 0, the only way for this inequality to hold is if:
\[ (a-b)^2 = 0 \]
which implies:
\[ a - b = 0 \quad \Rightarrow \quad a = b \]

### Conclusion
Thus, the condition \( \sec^2 \theta = \frac{4ab}{(a+b)^2} \) holds true if and only if:
\[ a = b \quad \text{and} \quad a, b \neq 0 \]

### Final Answer:
The solution is \( a = b \) where \( a \) and \( b \) are not equal to zero.

---