If `(a+b)^2/(4ab)=sin^2theta`, then

Options

A. (A)a=b
B. (B)`a ne 0, b ne 0`
C. (C)`a gt b`
D. (D)`a lt b`

Answer

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

### Step 1: Rewrite the equation
Start with the given equation:
\[ \frac{(a+b)^2}{4ab} = \sin^2 \theta \]

### Step 2: Cross-multiply
Cross-multiply to eliminate the fraction:
\[ (a+b)^2 = 4ab \sin^2 \theta \]

### Step 3: Expand the left side
Expand \((a+b)^2\):
\[ a^2 + 2ab + b^2 = 4ab \sin^2 \theta \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ a^2 + b^2 + 2ab - 4ab \sin^2 \theta = 0 \]
This simplifies to:
\[ a^2 + b^2 + 2ab(1 - 2\sin^2 \theta) = 0 \]

### Step 5: Use the identity for sine
Recall that \(1 - 2\sin^2 \theta = \cos 2\theta\):
\[ a^2 + b^2 + 2ab \cos 2\theta = 0 \]

### Step 6: Analyze the quadratic equation
This can be treated as a quadratic in terms of \(a\):
\[ a^2 + 2ab \cos 2\theta + b^2 = 0 \]

### Step 7: Calculate the discriminant
The discriminant \(\Delta\) of this quadratic must be non-negative for \(a\) to have real solutions:
\[ \Delta = (2b \cos 2\theta)^2 - 4(b^2) = 4b^2(\cos^2 2\theta - 1) \]
Since \(\cos^2 2\theta - 1 = -\sin^2 2\theta\), we have:
\[ \Delta = -4b^2 \sin^2 2\theta \]

### Step 8: Determine conditions for real solutions
For the discriminant to be non-negative:
\[ -4b^2 \sin^2 2\theta \geq 0 \]
This implies:
\[ \sin^2 2\theta = 0 \quad \Rightarrow \quad 2\theta = n\pi \quad \Rightarrow \quad \theta = \frac{n\pi}{2} \]
where \(n\) is an integer.

### Step 9: Conclusion
The only case where the quadratic has real solutions is when \(a = b\) or when \(\sin 2\theta = 0\).

### Final Result
Thus, we conclude that \(a = b\).

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