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In a $\triangle A B C$, the sides $a, b$ and $c$ are in $A$. P. Then $\left(\tan \frac{A}{2}+\tan \frac{C}{2}\right): \cot \frac{B}{2}$ is equal to
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The correct answer is:
$2: 3$
$\left(\tan \frac{A}{2}+\tan \frac{C}{2}\right): \cot \frac{B}{2}$
$=\left[\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}\right]: \sqrt{\frac{s(s-b)}{(s-c)(s-a)}}$
$=\frac{(s-c)+(s-c)}{\sqrt{s}}: \sqrt{s}$
$=2 s-(a+c): s$
$\Rightarrow \mathbf{b}: \frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}$
$\Rightarrow 2 b: a+b+c=2 b: 3 b$ $[\because \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P. $\therefore 2 \mathrm{~b}=\mathrm{a}+\mathrm{c}]$
$=2: 3$
$=\left[\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}\right]: \sqrt{\frac{s(s-b)}{(s-c)(s-a)}}$
$=\frac{(s-c)+(s-c)}{\sqrt{s}}: \sqrt{s}$
$=2 s-(a+c): s$
$\Rightarrow \mathbf{b}: \frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}$
$\Rightarrow 2 b: a+b+c=2 b: 3 b$ $[\because \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P. $\therefore 2 \mathrm{~b}=\mathrm{a}+\mathrm{c}]$
$=2: 3$
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