In $\mathrm{△}ABC\text{, if }BC\text{ is unity, }\sin \frac{A}{2}=x_1,\sin \frac{B}{2}=x_2,\cos \frac{A}{2}=x_3$ and $\cos \frac{B}{2}=x_4\text{ with }{\left(\frac{x_1}{x_2}\right)}^{2007}-{\left(\frac{x_3}{x_4}\right)}^{2006}=0$ then the of AC is………..

In given $\mathrm{△}\mathrm{ABC}\text{ both }\frac{\mathrm{A}}{2}\text{ and }\frac{\mathrm{B}}{2}\text{ lie strictly between }\left(0,\frac{\pi }{2}\right)$

and $\sin x$ is always increasing in $\left(0,\frac{\pi }{2}\right)$ whereas $\cos x$ is

always decreasing throughout $\left(0,\frac{\pi }{2}\right)$

So, if $\frac{A}{2}>\frac{B}{2}$

$\Rightarrow \sin \frac{A}{2}>\sin \frac{B}{2}$

or $x_1>x_2\text{ and }\frac{1}{x_3}>\frac{1}{x_4}$

So, $x_1^{2007}{x}_4^{2006}=x_2^{2007}{x}_3^{2006}$ is not valid.

Similarly for $\frac{A}{2}<\frac{B}{2}$

$\begin{array}{l}\Rightarrow \sin \frac{A}{2}<\sin \frac{B}{2}\\ \Rightarrow x_1<x_2\end{array}$

and $\frac{1}{x_3}<\frac{1}{x_4}$ again equality is not possible

Therefore, only possible case is when $\frac{A}{2}=\frac{B}{2}$

$\Rightarrow x_1=x_2\text{ and }\frac{1}{x_3}=\frac{1}{x_4}$

Hence, in that case $∆ABC$ is isosceles with $\mathrm{\angle }\mathrm{ABC}=\mathrm{\angle }\mathrm{CAB}$

$\Rightarrow \mathrm{BC}=\mathrm{AC}=1\text{ unit }$