Consider a $\triangle ABC ,$ right-angled at $B$.

$\tan A=\frac{\text { Side opposite to } \angle A }{\text { Side adjacent to } \angle A }$

$=\frac{12}{5}$

But $\frac{12}{5}>1$

$\therefore \tan A>1$

So, tan $A<1$ is not always true.

Hence, the given statement is false.

$(ii)$ $\sec A=\frac{12}{5}$

$\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle A }=\frac{12}{5}$

$\frac{A C}{A B}=\frac{12}{5}$

Let $AC$ be $12 k , AB$ will be $5 k ,$ where $k$ is a positive integer.

Applying Pythagoras theorem in $\triangle ABC ,$ we obtain

$AC ^{2}= AB ^{2}+ BC ^{2}$

$(12 k)^{2}=(5 k)^{2}+ BC ^{2}$

$144 k^{2}=25 k^{2}+B C^{2}$

$BC ^{2}=119 k ^{2}$

$BC =10.9 k$

It can be observed that for given two sides $AC =12 k$ and $AB =5 k$,

BC should be such that,

$AC – AB < BC < AC + AB$

$12 k-5 k< BC <12 k+5 k$

$7 k< BC <17 k$

However, $BC =10.9 k$. Clearly, such a triangle is possible and hence, such value of $\sec A$ is Possible.

Hence,the given statement is false.