Mathematics - Binomial Theorem Question with Solution | TestHub

$T_{r+1}={}^{16}C_r{\left(\frac{x}{\cos \theta }\right)}^{16-r}{\left(\frac{1}{x\sin \theta }\right)}^r$
for$r=8$term is free from$\text{'}x\text{'}$
$T_9={}^{16}C_8\frac{1}{{\sin }^8\theta {\cos }^8\theta }$
$T_9={}^{16}C_8\frac{2^8}{{\left(\sin 2\theta \right)}^8}$
In$\theta \in \left[\frac{\pi }{8},\frac{\pi }{4}\right],l_1={}^{16}C_8{2}^8$($\because$Minimum value of$l_1$at$\theta =\frac{\mathrm{\pi }}{4}$)
In$\theta \in \left[\frac{\pi }{16},\frac{\pi }{8}\right],l_2={}^{16}C_8\frac{2^8}{{\left(\frac{1}{\sqrt{2}}\right)}^8}={}^{16}C_8·2^8·2^4$($\because$Minimum value of$l_2$at$\theta =\frac{\pi }{8}$)
$\frac{l_2}{l_1}=\frac{{}^{16}C_8.2^8{2}^4}{{}^{16}C_8.2^8}=16$