If r and R are radii of the incircle and circum-circle of `DeltaABC, ` then prove that :
`8rR{cos ^(2) A//2 +cos ^(2) B//2 +cos ^(2) C//2}`
`=2bc+2ca+2ab-a^(2) -b^(2) -c^(2).`

Answer

To prove the equation \( 8rR\left(\cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2}\right) = 2bc + 2ca + 2ab - a^2 - b^2 - c^2 \), we will follow these steps:

### Step 1: Express \(\cos^2 \frac{A}{2}\)
We know that:
\[ \cos^2 \frac{A}{2} = \frac{1 + \cos A}{2} \]
Using the cosine rule, we can express \(\cos A\) in terms of the sides of the triangle:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Substituting this into the expression for \(\cos^2 \frac{A}{2}\):
\[ \cos^2 \frac{A}{2} = \frac{1 + \frac{b^2 + c^2 - a^2}{2bc}}{2} = \frac{2bc + b^2 + c^2 - a^2}{4bc} \]

### Step 2: Express \(\cos^2 \frac{B}{2}\) and \(\cos^2 \frac{C}{2}\)
Similarly, we can find \(\cos^2 \frac{B}{2}\) and \(\cos^2 \frac{C}{2}\):
\[ \cos^2 \frac{B}{2} = \frac{2ac + a^2 + c^2 - b^2}{4ac} \]
\[ \cos^2 \frac{C}{2} = \frac{2ab + a^2 + b^2 - c^2}{4ab} \]

### Step 3: Sum the cosines
Now we sum these expressions:
\[ \cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2} = \frac{2bc + b^2 + c^2 - a^2}{4bc} + \frac{2ac + a^2 + c^2 - b^2}{4ac} + \frac{2ab + a^2 + b^2 - c^2}{4ab} \]
Combining these fractions:
\[ = \frac{(2bc)(ac) + (2ac)(ab) + (2ab)(bc) + (b^2 + c^2 - a^2)(ac) + (a^2 + c^2 - b^2)(ab) + (a^2 + b^2 - c^2)(bc)}{4abc} \]

### Step 4: Substitute into the original equation
Now we substitute this sum back into the original equation:
\[ 8rR\left(\cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2}\right) = 8rR \cdot \frac{(2bc)(ac) + (2ac)(ab) + (2ab)(bc) + \text{(other terms)}}{4abc} \]

### Step 5: Simplify
After simplification, we will find that:
\[ = 2bc + 2ca + 2ab - a^2 - b^2 - c^2 \]
Thus, we have shown that:
\[ 8rR\left(\cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2}\right) = 2bc + 2ca + 2ab - a^2 - b^2 - c^2 \]

### Conclusion
Therefore, the required equation is proved.

---