In the interval $(0, \pi / 2)$, the function $f(x)=\tan ^n x+\cot ^n x$ attains
Answer & explanation
Correct answer: option 1
We have,
$f(x) =\tan ^n x+\cot ^n x$
$\Rightarrow f(x) =\left(\tan ^{n / 2} x-\cot ^{n / 2} x\right)^2+2 \geq 2$
Clearly, f(x) attains the minimum value 2 when $\tan ^{n / 2} x-\cot ^{n / 2} x=0$ i.e. $x=\frac{\pi}{4}$