Question

If the roots of the equation $x^2-kx+m=0$ is ${\tan }^{2}x$ and ${\cot }^{2}x,$ then the minimum possible value of $k+m$ is

Answer 1

Roots are ${\tan }^{2}x$ and ${\cot }^{2}x$
We know that ${\tan }^{2}x+{\cot }^{2}x=\frac{-b}{a}=k$
Also, ${\tan }^{2}x\cdot {\cot }^{2}x=\frac{c}{a}=m$
$\Rightarrow {\tan }^{2}x\cdot {\cot }^{2}x=1=m$

${\tan }^{2}x+\frac{1}{{\tan }^{2}x}=k$
Let ${\tan }^{2}x=X$
$\Rightarrow X+\frac{1}{X}=k$
We know that the minimum value of $X+\frac{1}{X}$ is $2$, where $X>0$
$\therefore$ Minimum value of $k+m$ is $2+1=3$