In a `DeltaABC`, if `4 cos A cos B + sin 2A+sin 2B+sin 2C=4`, then `DeltaABC` is

Options

A. right angle but not issosceles
B. isosceles but not right angled
C. right angle isosceles
D. obtuse angled

Answer

To solve the problem, we need to analyze the given equation in the context of triangle \( \Delta ABC \):

**Given:**
\[ 4 \cos A \cos B + \sin 2A + \sin 2B + \sin 2C = 4 \]

### Step 1: Rewrite the equation
We can start by rewriting the equation:
\[ 4 \cos A \cos B + \sin 2A + \sin 2B + \sin 2C = 4 \]

### Step 2: Use the identity for sine
Recall that \( \sin 2A = 2 \sin A \cos A \), \( \sin 2B = 2 \sin B \cos B \), and \( \sin 2C = 2 \sin C \cos C \). However, we will use the sine addition formula later.

### Step 3: Simplifying the equation
We can rewrite the equation as:
\[ 4 \cos A \cos B + \sin 2A + \sin 2B + \sin 2C = 4 \]
This can be rearranged to:
\[ 4 \cos A \cos B + \sin 2A + \sin 2B + \sin 2C - 4 = 0 \]

### Step 4: Analyze the maximum values
The maximum value of \( \sin 2A, \sin 2B, \sin 2C \) is 1. Therefore, the maximum value of \( \sin 2A + \sin 2B + \sin 2C \) can be at most 3. Thus, we have:
\[ 4 \cos A \cos B + \text{(maximum of 3)} \leq 4 \]
This implies:
\[ 4 \cos A \cos B \leq 1 \implies \cos A \cos B \leq \frac{1}{4} \]

### Step 5: Consider the equality case
The equality holds when \( \cos A \cos B = 0 \). This occurs when either \( A = 90^\circ \) or \( B = 90^\circ \).

### Step 6: Use the sine rule
If \( C = 90^\circ \), then by the sine rule, we have:
\[ \sin C = 1 \]
This leads us to conclude that \( C \) must be \( 90^\circ \).

### Step 7: Analyze the triangle
Since \( C = 90^\circ \), and we also have \( A = B \) (from the cosine condition), we conclude that \( \Delta ABC \) is an isosceles right triangle.

### Conclusion
Thus, we conclude that \( \Delta ABC \) is an **isosceles right triangle**.

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