In a , then is equal to

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In
a $\bigtriangleup ABC,\ if\ \frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13}$ , then $\tan^{2}\frac{A}{2}$ is
equal to

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\frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13} = \frac{3s - (a + b + c)}{11 + 12 + 13} = \frac{s}{36}

Now

\tan^{2}\left( \frac{A}{2} \right) = \frac{(s - b)(s - c)}{s(s - a)} = \frac{(12)(13)}{(11)(36)} = \frac{13}{33}

Source: Grass-G / M-24 Mathematics Paper 31 Dec.docx

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