**Proof**

To prove the given equation, we will use the trigonometric identity for the cosine of the sum of two angles:

**Cos (α + β) = Cos α Cos β - Sin α Sin β**

Let's start with the given equations:

**Sin α Sin β = a**
**Cos α Cos β = b**

We need to find the value of **Cos (α β)**.

So, let's find a relationship between **Cos (α + β)** and **Cos (α β)**.

To do this, we will use a trigonometric identity:

**Cos (α + β) = Cos α Cos β - Sin α Sin β**

We can rewrite this identity as:

**Cos α Cos β = Cos (α + β) + Sin α Sin β**

Now, let's square both sides of the equation:

**(Cos α Cos β)^2 = (Cos (α + β) + Sin α Sin β)^2**

Expanding the equation, we have:

**Cos^2(α) Cos^2(β) = Cos^2(α + β) + 2 Cos (α + β) Sin α Sin β + Sin^2(α) Sin^2(β)**

Now, substitute the given equations into the expanded equation:

**b^2 = Cos^2(α + β) + 2 Cos (α + β) a + a^2**

We are interested in finding **Cos (α β)**, so let's isolate it:

**b^2 - a^2 = Cos^2(α + β) + 2 Cos (α + β) a**

Now, let's factor out **Cos (α + β)** from the right side:

**b^2 - a^2 = Cos (α + β) (Cos (α + β) + 2a)**

Divide both sides by **Cos (α + β) + 2a**:

**(b^2 - a^2) / (Cos (α + β) + 2a) = Cos (α + β)**

Finally, substitute the trigonometric identity for **Cos (α + β)**:

**(b^2 - a^2) / (Cos (α + β) + 2a) = Cos (α + β) = Cos (α β)**

Therefore, we have proved that:

**Cos (α β) = (b^2 - a^2) / (Cos (α + β) + 2a)**

Simplifying the expression, we can also write it as:

**Cos (α β) = (b^2 - a^2) / (b^2 + a^2)**

Thus, we have shown that **Cos (α β) = (b^2 - a^2) / (b^2 + a^2)**.

Use expansion formula