Verify this!

In $\big(0;\frac{\pi}{6}\big)$ both $sinx$ and $tanx$ increase. So now consider the $LHS$ $LHS < sin\big(\frac{\pi}{6}\big)+tan^2\big(\frac{\pi}{6}\big)+sin^3\big(\frac{\pi}{6}\big)+tan^4\big(\frac{\pi}{6}\big)+...+sin^{2n-1}\big(\frac{\pi}{6}\big)+tan^{2n}\big(\frac{\pi}{6}\big)$ The above $RHS$ equals to $\frac{1}{2}+\bigg(\frac{1}{\sqrt{3}}\bigg)^2+...+\bigg(\frac{1}{2}\bigg)^{2n-1}+\bigg(\frac{1}{\sqrt{3}}\bigg)^{2n}$ Now we consider these following sum $A=\frac{1}{2}+\big(\frac{1}{2}\big)^3+...+\big(\frac{1}{2}\big)^{2n-1}$ and $B=\big(\frac{1}{\sqrt{3}}\big)^2+\big(\frac{1}{\sqrt{3}}\big)^4+...+\big(\frac{1}{\sqrt{3}}\big)^{2n}$ . We see that $\bigg(\frac{1}{2}\bigg)^2A=\bigg(\frac{1}{2}\bigg)^3+ \bigg(\frac{1}{2}\bigg)^5...+\bigg(\frac{1}{2}\bigg)^{2n+1}$ $\bigg(\frac{1}{\sqrt{3}}\bigg)^2B=\bigg(\frac{1}{\sqrt{3}}\bigg)^4+\bigg(\frac{1}{\sqrt{3}}\bigg)^6...+\bigg(\frac{1}{\sqrt{3}}\bigg)^{2n+2}$ Therefore $A-\frac{1}{4}A=\frac{1}{2}-\bigg(\frac{1}{2}\bigg)^{2n+1}\Rightarrow A=\frac{4}{3}\bigg[\frac{1}{2}-\bigg(\frac{1}{2}\bigg)^{2n+1}\bigg]<\frac{4}{3}.\frac{1}{2}=\frac{2}{3}$ $B-\frac{1}{3}B=\frac{1}{3}-\bigg(\frac{1}{\sqrt{3}}\bigg)^{2n+2}\Rightarrow B=\frac{3}{2}\bigg[\frac{1}{3}-\bigg(\frac{1}{\sqrt{3}}\bigg)^{2n+2}\bigg]<\frac{3}{2}.\frac{1}{2}=\frac{1}{2}$ So $LHS<A+B<\frac{2}{3}+\frac{1}{2}=\frac{7}{6}<\sqrt{2}$ , the correct choice is "False for all $n$ ".