Given: $\tan (A+B-C)=1$
$\implies \tan(A+B-C)=\tan 45^\circ$
$\implies (A+B-C)=45^\circ$ …...(i)
In same way $\sec(B+C-A)=2$
$\implies \sec(B+C-A)=\sec 60^\circ$
$\implies (B+C-A)=60^\circ$……..(ii)
from equation (i) & (ii)
$2B=105^\circ\implies B=52.5^\circ$
For any triangle
$\because \angle A+\angle B+\angle C=180^\circ$
$\implies A+B=180^\circ-C$…..(iii)
from equation (i)&(iii),
$\implies 180^\circ-C-C=45^\circ$
$\implies 2C=135^\circ \implies 67.5^\circ$
now form equation (iii) we get:
$\angle A+52.5^\circ=180^\circ-67.5^\circ$
$\angle A=60^\circ$
Option $(A)$ is correct.