In the expansion of ${\left( {{x \over {\cos \theta }} + {1 \over {x\sin \theta }}} \right)^{16}}$, if ${\ell _1}$ is the least value of the term independent of x when ${\pi \over 8} \le \theta \le {\pi \over 4}$ and ${\ell _2}$ is the least value of the term independent of x when ${\pi \over {16}} \le \theta \le {\pi \over 8}$, then the ratio ${\ell _2}$ : ${\ell _1}$ is equal to :

T_{r + 1} = ¹⁶C_r$${\left( {{x \over {\cos \theta }}} \right)^{16 - r}}{\left( {{1 \over {x\sin \theta }}} \right)^r}$$ <br><br>= ¹⁶C_r$${\left( x \right)^{16 - 2r}} \times {1 \over {{{\left( {\cos \theta } \right)}^{16 - r}}{{\left( {\sin \theta } \right)}^r}}}$$ <br><br>term is independent of x when <br><br>$\therefore$ 16 – 2r = 0 <br><br>$\Rightarrow$ r = 8 <br><br>T₉ = ¹⁶C₈ $\times$ ${1 \over {{{\cos }^8}\theta {{\sin }^8}\theta }}$ <br><br>= ¹⁶C₈ $\times$ ${{{2^8}} \over {{{\left( {\sin 2\theta } \right)}^8}}}$ <br><br>If $\theta \in \left[ {{\pi \over 8},{\pi \over 4}} \right]$ then $2\theta \in \left[ {{\pi \over 4},{\pi \over 2}} \right]$ <br><br>In the range $\left[ {{\pi \over 4},{\pi \over 2}} \right]$, sin 2$\theta$ is increasing. <br><br>And value of T₉ is least when sin 2$\theta$ is maximum. <br><br>And sin 2$\theta$ is maximum in the range $\left[ {{\pi \over 4},{\pi \over 2}} \right]$ <br>when 2$\theta$ = ${{\pi \over 2}}$ <br><br>$\therefore$ ${l_1}$ = ¹⁶C₈ $\times$ 2⁸ <br><br>AgainIf $\theta \in \left[ {{\pi \over {16}},{\pi \over 8}} \right]$ then $2\theta \in \left[ {{\pi \over 8},{\pi \over 4}} \right]$ <br><br>In the range $\left[ {{\pi \over 8},{\pi \over 4}} \right]$, sin 2$\theta$ is increasing. <br><br>And value of T₉ is least when sin 2$\theta$ is maximum. <br><br>And sin 2$\theta$ is maximum in the range $\left[ {{\pi \over 8},{\pi \over 4}} \right]$ <br>when 2$\theta$ = ${{\pi \over 4}}$ <br><br>$\therefore$ ${l_2}$ = ¹⁶C₈ $\times$ ${{{2^8}} \over {{{\left( {{1 \over {\sqrt 2 }}} \right)}^8}}}$ <br><br> = ¹⁶C₈ $\times$ ${{2^8}{2^4}}$ <br><br>$\therefore$ ${{{l_2}} \over {{l_1}}}$ = ${{{2^4}} \over 1}$ = ${{16} \over 1}$