Answer
$2k\pi$
Work Step by Step
$\tan \theta+1=\sec \theta$
$\frac{\sin\theta}{\cos\theta}+1=\frac{1}{\cos \theta}$
Multiply both sides by $\cos \theta$:
$\sin \theta+\cos \theta=1$
$\sin(2\times\frac{\theta}{2})+\cos(2\times\frac{\theta}{2})=1$
$2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+1-2\sin^2\frac{\theta}{2}=1$
$2\sin\frac{\theta}{2}\cos \frac{\theta}{2}-2\sin^2\frac{\theta}{2}=0$
$2\sin\frac{\theta}{2}(\cos \frac{\theta}{2}-\sin \frac{\theta}{2})=0$
If $2\sin \frac{\theta}{2}=0$, then $\sin \frac{\theta}{2}=0$. This means that $\frac{\theta}{2}=k\pi$, so $\theta=2k\pi$.
If $\cos\frac{\theta}{2}-\sin\frac{\theta}{2}=0$, then $\cos \frac{\theta}{2}=\sin \frac{\theta}{2}$. This means that $\frac{\theta}{2}=\frac{\pi}{4}+k\pi$, so $\theta=\frac{\pi}{2}+2k\pi$. However, $\theta=\frac{\pi}{2}+2k\pi$ would make $\tan \theta$ and $\sec \theta$ undefined in the original equation, so we reject this solution, and the only solution is $\theta=2k\pi$.
