If $a > 1$, taking logs of both sides we obtain the equivalent inequality
$$\sin x + \cos x \cdot \log_a (a+1) \ge 1.$$
For $x \in [0, \pi/2]$, we have $\sin x \ge \sin^2 x$ and $\cos x \ge \cos^2 x$. Because $\log_a (a+1) > 1$, we obtain
$$\sin x + \cos x \cdot \log_a (a+1) \ge \sin^2 x + \cos^2 x = 1.$$
The inequality clearly holds true for $a = 1$.

Finally, if $a \in (0, 1)$, the inequality is equivalent to
$$\sin x + \cos x \cdot \log_a (a+1) \le 1, \quad \forall x \in [0, \pi/2],$$
which is obvious, since $\sin x \le 1$, $\cos x \ge 0$ and $\log_a (a+1) < 0$.