$\sec A=(\frac{12}{5}$) for some value of angle $A$. Is it True/False?

The correct option is B (True)

Given $\sec A=\frac{12}{5}=\frac{\text{Hypotenuse}}{\text{Base}}$
Let $\Delta MNC$ be a right angled triangle such that $MC=12k$ and $NC=5k$ as shown in figure.

By Pythagoras theorem, we get
$M{C}^2=M{N}^2+N{C}^2$
$\Rightarrow (12k{)}^2=M{N}^2+(5k{)}^2$
$\Rightarrow 144k^2=M{N}^2+25k^2$
$\Rightarrow M{N}^2=144k^2-25k^2$
$\Rightarrow M{N}^2=119k^2$ which exists for at least some real number $k$.

Therefore, $\sec A=\frac{12}{5}$ exist for some value of angle $A$.
Hence, the statement is true.