How to prove this inequality involving $\tanh$?

The complexity of your formulae is distracting. Let \begin{gather*} r=\frac{\sin{\alpha}}{\sin{\theta}} \\ s=\frac{\sin{(\theta-\alpha)}}{\sin{\theta}} \end{gather*} and $T(x)=(\tanh{x^{-1}})^{-1}$. Then we want to show the Jensen-type inequality $$T(ra^{-1}+sb^{-1})\leq rT(a^{-1})+sT(b^{-1})$$

It is straightforward to verify that $T$ is convex. So, if $r+s=1$, then Jensen solves the problem immediately.

Otherwise, note that concavity implies $T(\lambda x)\leq\lambda T(x)$ for all $\lambda,x>0$. So let $\lambda=r+s$ and $(\tilde{r},\tilde{s})=\lambda^{-1}\cdot(r,s)$; then \begin{align*} T(ra^{-1}+sb^{-1})&=T(\lambda\cdot(\tilde{r}a^{-1}+\tilde{s}b^{-1})) \\ &\leq\lambda T(\tilde{r}a^{-1}+\tilde{s}b^{-1}) \\ &\leq\lambda\cdot(\tilde{r}T(a^{-1})+\tilde{s}T(b^{-1})) \\ &=rT(a^{-1})+sT(b^{-1}) \end{align*}