10. If real numbers $b, c$ satisfy $b^{2}+c^{2}=1$, and
$$
f(x)=a x+b \sin x+c \cos x
$$

has two perpendicular tangent lines on its graph, then the range of values for $a$ is $\qquad$.
10. $\{0\}$.

From the problem, we have
$$
\begin{aligned}
f(x) & =a x+\sqrt{b^{2}+c^{2}} \sin (x+\varphi) \\
\therefore \quad & =a x+\sin (x+\varphi) \\
\Rightarrow & f^{\prime}(x)=a+\cos (x+\varphi)
\end{aligned}
$$

where, $\sin \varphi=\frac{c}{\sqrt{b^{2}+c^{2}}}, \cos \varphi=\frac{b}{\sqrt{b^{2}+c^{2}}}$.
If the graph of $f(x)$ has two perpendicular tangents, then there exist $x_{1} 、 x_{2}$ such that
$$
\left[a+\cos \left(x_{1}+\varphi\right)\right]\left[a+\cos \left(x_{2}+\varphi\right)\right]=-1 \text {. }
$$

Let $\cos \left(x_{i}+\varphi\right)=u_{i}(i=1,2)$.
Thus, $-1 \leqslant u_{i} \leqslant 1$.
From equation (1), we get
$$
\begin{array}{l}
a^{2}+\left(u_{1}+u_{2}\right) a+u_{1} u_{2}+1=0 . \\
\text { Therefore, } \Delta=\left(u_{1}+u_{2}\right)^{2}-4 u_{1} u_{2}-4 \\
=\left(u_{1}-u_{2}\right)^{2}-4 \geqslant 0 \\
\Rightarrow\left|u_{1}-u_{2}\right| \geqslant 2 \text {. } \\
\text { Also, }-1 \leqslant u_{i} \leqslant 1 \text {, so } \\
u_{1}=-u_{2}, u_{1} u_{2}=-1 .
\end{array}
$$

Therefore, $a=0$, which means the range of $a$ is $\{0\}$.