$\frac{cot (A / 2) + cot (B / 2) + cot (C / 2)}{cot A + cot B + cot C} =$

\frac{(a + b + c)^{2}}{a^{2} + b^{2} + c^{2}}

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\frac{a^{2} + b^{2} + c^{2}}{(a + b + c)^{2}}

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s

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Δ

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Solution

The correct option is A $\frac{(a + b + c)^{2}}{a^{2} + b^{2} + c^{2}}$
$cot \frac{A}{2} = \frac{s (s - a)}{Δ}$

$cot \frac{B}{2} = \frac{s (s - b)}{Δ}$

$cot \frac{C}{2} = \frac{s (s - c)}{Δ}$

$cot A = \frac{b^{2} + c^{2} - a^{2}}{4 Δ}$

$= \frac{\frac{s (s - a) + s (s - b) + s (s - c)}{Δ}}{\frac{a^{2} + b^{2} + c^{2}}{4 Δ}}$

$= \frac{4 s (3 s - a - b - c)}{a^{2} + b^{2} + c^{2}}$

$= \frac{4 s (\frac{a + b + c}{2})}{a^{2} + b^{2} + c^{2}}$

$= \frac{\frac{(a + b + c)}{2} \frac{(a + b + c)}{2} \times 4}{a^{2} + b^{2} + c^{2}}$

$= \frac{(a + b + c)^{2}}{a^{2} + b^{2} + c^{2} .}$

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Similar questions

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

a² + b² + c² - ab - bc - ca equals:

\frac{a^{2} - {(b - c)}^{2}}{{(a + c)}^{2} - b^{2}} + \frac{b^{2} - {(a - c)}^{2}}{{(a + b)}^{2} - c^{2}} + \frac{c^{2} - {(a - b)}^{2}}{{(b + c)}^{2} - a^{2}}

\frac{a^{2} - b^{2} - c^{2}}{(a - b) (a - c)} + \frac{b^{2} - c^{2} - a^{2}}{(b - c) (b - a)} + \frac{c^{2} - a^{2} - b^{2}}{(c - a) (c - b)}

In a $△ A B C, 2 c a s i n (\frac{A - B + C}{2})$ is equal to

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