Question

In a ∆ABC
prove that

cot(A/2) + cot(B/2) + Cot(C/2) = (S²/∆)
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Answer 1

$\huge{\underline{\mathfrak{\blue{Solution}}}}$

$\sqrt{\dfrac{s(s-b)}{(s-a)(s-c)}} + \sqrt{\dfrac{s(s-c)}{(s-b)(s-a)}} +\sqrt{\dfrac{s(s-c)}{(s-b)(s-a)}} \\ \\ \implies \dfrac{s(s-a)}{ \triangle} +\dfrac{s(s-b)}{ \triangle} +\dfrac{s(s-c)}{ \triangle} \\ \\ \implies \dfrac{3s^2 -s(a+b+c)}{\triangle} \\ \\ \implies \dfrac{3s^2 - 2s^2 }{\triangle} \\ \\ \huge\boxed{\dfrac{s^2}{\triangle}}$

Answer 2

$<marquee> ☺ ❤ HeYa maTes ❤ ☺ </marquee>$

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$\huge\underline\mathcal\red{QueStion}$

In a ∆ABC

prove that

cot(A/2) + cot(B/2) + Cot(C/2) = (S²/∆)

$\huge\underline\mathfrak\red{Answer!!!!!}$

refer to the attachment..