Question:

Step 1: Use the sum-to-product formula for $\cos A + \cos B$. $$\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$$ Since $A+B+C=\pi$, we have $\frac{A+B}{2} = \frac{\pi - C}{2}$, so $\cos\frac{A+B}{2} = \sin\frac{C}{2}$. $$\cos A + \cos B = 2\sin\frac{C}{2}\cos\frac{A-B}{2}$$ Step 2: Set equal to the right side and simplify. $$2\sin\frac{C}{2}\cos\frac{A-B}{2} = 4\sin^2\frac{C}{2}$$ $$\cos\frac{A-B}{2} = 2\sin\frac{C}{2} = 2\cos\frac{A+B}{2}$$ $$\cos\frac{A-B}{2} = 2\cos\frac{A+B}{2}$$ Step 3: Expand using sum-to-product. $$\cos\frac{A-B}{2} - 2\cos\frac{A+B}{2} = 0$$ Using $\cos P - \cos Q$ identities: $$\cos\frac{A-B}{2} - \cos\frac{A+B}{2} = 2\sin\frac{A}{2}\sin\frac{B}{2}$$ So: $2\sin\frac{A}{2}\sin\frac{B}{2} = \cos\frac{A+B}{2}$ This gives: $2\sin\frac{A}{2}\sin\frac{B}{2} = \cos\frac{A+B}{2}$. Step 4: Use the sine rule to check $a + b = 2c$. By the sine rule, $a = 2R\sin A$, $b = 2R\sin B$, $c = 2R\sin C$. The condition $\cos\frac{A-B}{2} = 2\cos\frac{A+B}{2}$ leads to: $$\sin A + \sin B = 2\sin C \Rightarrow a + b = 2c. \checkmark$$ Step 5: Check if $a, b, c$ are in H.P. H.P. requires $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ in A.P., i.e., $b(a+c) = 2ac$. This is not implied by $a+b=2c$ in general, so option (2) is false. Step 6: Check option (3): $\tan\frac{A}{2}, \tan\frac{C}{2}, \tan\frac{B}{2}$ in A.P. Using $\tan\frac{A}{2} = \frac{r}{s-a}$, $\tan\frac{B}{2} = \frac{r}{s-b}$, $\tan\frac{C}{2} = \frac{r}{s-c}$, where $s = \frac{a+b+c}{2}$. For A.P.: $2\tan\frac{C}{2} = \tan\frac{A}{2} + \tan\frac{B}{2}$ $$\frac{2r}{s-c} = \frac{r}{s-a} + \frac{r}{s-b} = \frac{r(2s-a-b)}{(s-a)(s-b)}$$ Since $a+b=2c$, $2s = a+b+c = 2c+c = 3c$, so $s = \frac{3c}{2}$. $s-c = \frac{c}{2}$, $s-a = \frac{3c}{2}-a$, $s-b = \frac{3c}{2}-b$. $2s-a-b = c$. $$\frac{2}{s-c} = \frac{c}{(s-a)(s-b)} \Rightarrow 2(s-a)(s-b) = c(s-c) = c\cdot\frac{c}{2} = \frac{c^2}{2}$$ This is consistent, confirming option (3). ✓