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Q.
The minimum value of the function $f(x)=\frac{\tan \left(x+\frac{\pi}{6}\right)}{\tan x}$ is
Application of Derivatives
Solution:
$ \mathrm{f}(\mathrm{x})$ has a period equal to $\pi \&$ can take values $(-\infty, \infty) \Rightarrow 3$ is the local minimum value.
$y=\frac{2 \sin \left(x+\frac{\pi}{6}\right) \cos x}{2 \sin x \cos \left(x+\frac{\pi}{6}\right)}=\frac{\sin \left(2 x+\frac{\pi}{6}\right)+\sin \frac{\pi}{6}}{\sin \left(2 x+\frac{\pi}{6}\right)-\sin \frac{\pi}{6}}$
$=1+\frac{1}{\sin \left(2 x+\frac{\pi}{6}\right)-\sin \frac{\pi}{6}}$
$\mathrm{y}$ is minimum if $2 \mathrm{x}+\frac{\pi}{6}=\frac{\pi}{2}$
$\left.\Rightarrow \mathrm{x}=\frac{\pi}{6} \Rightarrow \mathrm{y}_{\min }=1+2=3\right]$
