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In a triangle $A B C$, if $c=9, s=10$ and $\Delta=10 \sqrt{2}$ then $b\left[1+\sqrt{2} \tan \left(\frac{A-B}{2}\right)\right]=$
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The correct answer is:
$a\left[1-\sqrt{2} \tan \left(\frac{A-B}{2}\right)\right]$
For a triangle $A B C$, it is given that $c=9, s=10$ and $\Delta=10 \sqrt{2}$
$\therefore \quad \tan \frac{A-B}{2}=\frac{a-b}{a+b} \cot \frac{c}{2}=\frac{a-b}{a+b} \frac{s(s-c)}{\Delta}$
$=\frac{a-b}{a+b} \times \frac{10 \times 1}{10 \sqrt{2}}=\frac{1}{\sqrt{2}} \frac{a-b}{a+b}$
So, $b\left[1+\sqrt{2} \tan \frac{A-B}{2}\right]$
$=b\left[1+\left(\sqrt{2} \times \frac{1}{\sqrt{2}} \frac{a-b}{a+b}\right)\right]=b\left[\frac{a+b+a-b}{a+b}\right]$
$=b\left[\frac{2 a}{a+b}\right]=a\left[\frac{2 b}{a+b}\right]=a\left[\frac{(a+b)-(a-b)}{a+b}\right]$
$=a\left[1-\frac{a-b}{a+b}\right]=a\left[1-\sqrt{2} \tan \frac{A-B}{2}\right]$
$\therefore \quad \tan \frac{A-B}{2}=\frac{a-b}{a+b} \cot \frac{c}{2}=\frac{a-b}{a+b} \frac{s(s-c)}{\Delta}$
$=\frac{a-b}{a+b} \times \frac{10 \times 1}{10 \sqrt{2}}=\frac{1}{\sqrt{2}} \frac{a-b}{a+b}$
So, $b\left[1+\sqrt{2} \tan \frac{A-B}{2}\right]$
$=b\left[1+\left(\sqrt{2} \times \frac{1}{\sqrt{2}} \frac{a-b}{a+b}\right)\right]=b\left[\frac{a+b+a-b}{a+b}\right]$
$=b\left[\frac{2 a}{a+b}\right]=a\left[\frac{2 b}{a+b}\right]=a\left[\frac{(a+b)-(a-b)}{a+b}\right]$
$=a\left[1-\frac{a-b}{a+b}\right]=a\left[1-\sqrt{2} \tan \frac{A-B}{2}\right]$
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