If sin(A – B) = 0, where 0° ≤ A, B ≥ 90° then find the value of 2 sin A × sin B + 2 cos A × cos B

Option 4 : Both options 1 and 2 are correct

**Given:**

Value of sin(A – B) = 0

**Identity used:**

sin^{2 }A + cos^{2 }A = 1

**Calculation:**

As, sin(A – B ) = 0

⇒ sin(A – B ) = sin 0°

⇒ A – B = 0

⇒ A = B

We have to find the value of 2 sin A × sin B + 2 cos A × cos B

⇒ 2 sin A × sin B + 2 cos A × cos B

⇒ 2 × (sin A × sin A + cos A × cos A)

⇒ 2 × (sin^{2 }A + cos^{2 }A)

⇒ 2

Now 2 is also the even prime number less than 5.

**∴ Both options 1 and 2 are correct**.