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Q.
If in the expansion of $\left(\frac{1}{x}+x \tan x\right)^5$ the ratio of the 4 th term to the 2 nd is $\frac{2}{27} \pi^4$, then the value of $x$ is
Binomial Theorem
Solution:
$T _2={ }^5 C _1\left(\frac{1}{ x }\right)^4 \cdot x \tan x =\frac{5 \tan x }{ x ^3}, T _4={ }^5 C _3\left(\frac{1}{ x }\right)^2 \cdot x ^3 \tan ^3 x =10 x \tan ^3 x$
$\frac{ T _4}{ T _2}=2 x ^4 \tan ^2 x =\frac{2 \pi^4}{27}$ or $x ^4 \tan ^2 x =\left(\frac{\pi}{3}\right)^4(\sqrt{3})^2$
$\therefore x =\frac{\pi}{3}$