Question

In a triangle ABC with usual notations, if a, b, c are in arithmetic progression, then, $\tan \frac{A}{2} \cdot \tan \frac{C}{2} =$}

• A. $3$
• B. $\frac{1}{13}$
• C. $-3$
• D. $\frac{1}{3}$

Hint:

If $a, b, c$ are in A.P., then $s = 1.5b$. This shortcut makes many triangle property questions easier.

Answer 1

The Correct Option is D

Step 1: Concept
Use half-angle formulas: $\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $\tan \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.

Step 2: Meaning
Multiplying them: $\tan \frac{A}{2} \cdot \tan \frac{C}{2} = \frac{s-b}{s}$.

Step 3: Analysis
Since $a, b, c$ are in A.P., $2b = a + c$. The semi-perimeter $s = \frac{a+b+c}{2} = \frac{2b+b}{2} = \frac{3b}{2}$. Substituting $s$: $\frac{(3b/2) - b}{3b/2} = \frac{b/2}{3b/2} = \frac{1}{3}$.

Step 4: Conclusion
The product is $1/3$. Final Answer: (D)