Question

In \( \triangle ABC \), if \( \tan \frac{A}{2} + \tan \frac{C}{2} = \frac{b}{s} \), then \( \sin \left( \frac{A + C}{3} \right) = \)

• A. \( 1 \)
• B. \( \frac{\sqrt{3}}{2} \)
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( \frac{1}{2} \)

Hint:

Use the identity \( \tan \frac{A}{2} + \tan \frac{C}{2} = \frac{\sin(\frac{A+C}{2})}{\cos \frac{A}{2} \cos \frac{C}{2}} \). Simplify \( \sin(\frac{A+C}{2}) \) using \( A + B + C = \pi \). Consider a specific type of triangle (e.g., right-angled triangle) that satisfies the given condition to find the value of \( \sin \left( \frac{A + C}{3} \right) \).

Answer 1

The Correct Option is D

We know that in a triangle \( ABC \), \( A + B + C = \pi \), so \( \frac{A + C}{2} = \frac{\pi}{2} - \frac{B}{2} \).
Given \( \tan \frac{A}{2} + \tan \frac{C}{2} = \frac{b}{s} \).
We also know that \( \tan \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} \) and \( \tan \frac{C}{2} = \sqrt{\frac{(s - a)(s - b)}{s(s - c)}} \).
So, \( \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} + \sqrt{\frac{(s - a)(s - b)}{s(s - c)}} = \frac{b}{s} \) $$ \sqrt{\frac{s - b}{s}} \left( \sqrt{\frac{s - c}{s - a}} + \sqrt{\frac{s - a}{s - c}} \right) = \frac{b}{s} $$ $$ \sqrt{\frac{s - b}{s}} \left( \frac{(s - c) + (s - a)}{\sqrt{(s - a)(s - c)}} \right) = \frac{b}{s} $$ $$ \sqrt{\frac{s - b}{s}} \left( \frac{2s - (a + c)}{\sqrt{(s - a)(s - c)}} \right) = \frac{b}{s} $$ Since \( a + b + c = 2s \), \( a + c = 2s - b \).
$$ \sqrt{\frac{s - b}{s}} \left( \frac{2s - (2s - b)}{\sqrt{(s - a)(s - c)}} \right) = \frac{b}{s} $$ $$ \sqrt{\frac{s - b}{s}} \left( \frac{b}{\sqrt{(s - a)(s - c)}} \right) = \frac{b}{s} $$ If \( b \neq 0 \), we can divide by \( b \): $$ \frac{\sqrt{s - b}}{\sqrt{s} \sqrt{(s - a)(s - c)}} = \frac{1}{s} $$ $$ s \sqrt{s - b} = \sqrt{s (s - a)(s - c)} $$ Squaring both sides: $$ s^2 (s - b) = s (s - a)(s - c) $$ If \( s \neq 0 \): $$ s (s - b) = (s - a)(s - c) $$ $$ s^2 - sb = s^2 - sc - sa + ac $$ $$ sa + sc - ac = sb $$ Dividing by \( abc \): $$ \frac{s}{bc} + \frac{s}{ab} - \frac{s}{b} = \frac{s}{ac} $$ This does not seem straightforward.
Consider \( \tan \frac{A}{2} + \tan \frac{C}{2} = \frac{\sin(\frac{A}{2} + \frac{C}{2})}{\cos \frac{A}{2} \cos \frac{C}{2}} = \frac{\sin(\frac{\pi - B}{2})}{\cos \frac{A}{2} \cos \frac{C}{2}} = \frac{\cos \frac{B}{2}}{\cos \frac{A}{2} \cos \frac{C}{2}} = \frac{b}{s} \) We need to find \( \sin \left( \frac{A + C}{3} \right) = \sin \left( \frac{\pi - B}{3} \right) \).
If \( B = \frac{\pi}{3} \), then \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \).
\( \frac{\sqrt{3}/2}{\cos \frac{A}{2} \cos \frac{C}{2}} = \frac{b}{s} \).
If \( A = C \), then \( 2 \tan \frac{A}{2} = \frac{b}{s} \).
\( B = \pi - 2A \).
\( \sin \left( \frac{2A}{3} \right) \).
Consider the case when \( a = c \).
Then \( A = C \).
\( 2 \tan \frac{A}{2} = \frac{b}{s} \).
\( \sin \left( \frac{2A}{3} \right) \).
If \( b^2 = a^2 + c^2 - 2ac \cos B = 2a^2 (1 - \cos B) \).
If \( B = \frac{\pi}{2} \), \( \tan \frac{A}{2} + \tan \frac{C}{2} = \tan \frac{A}{2} + \cot \frac{A}{2} = \frac{\sin^2 \frac{A}{2} + \cos^2 \frac{A}{2}}{\sin \frac{A}{2} \cos \frac{A}{2}} = \frac{1}{\frac{1}{2} \sin A} = \frac{2}{\sin A} = \frac{b}{s} \).
\( \sin(\frac{A+C}{3}) = \sin(\frac{\pi/2}{3}) = \sin \frac{\pi}{6} = \frac{1}{2} \).