Let $a>0, b >0$ and let $0<\alpha<\theta<\pi$. Prove that $$\frac{1}{\tanh \left(\frac{1}{\frac{\sin (\alpha )}{a \sin (\theta )}+\frac{\sin (\theta -\alpha )}{b \sin (\theta )}}\right)}\leq \frac{\sin (\alpha )}{\tanh (a) \sin (\theta )}+\frac{\sin (\theta -\alpha )}{\tanh (b) \sin (\theta )}.$$
My attempts: Firstly, plugging randomly chosen values for $a,b,\alpha,\theta$ many times, this does indeed seem to be true.
If $a=b$, then I can prove that this is true using the identity $\tanh(x y) \leq x \tanh(y)$ for all $x \geq 1$ and $y > 0$. Indeed, when $a=b$ the inequality is equivalent to $$\tanh (a)\leq \left(\frac{\sin (\theta -\alpha )}{\sin (\theta )}+\frac{\sin (\alpha )}{\sin (\theta )}\right) \tanh \left(\frac{a}{\frac{\sin (\theta -\alpha )}{\sin (\theta )}+\frac{\sin (\alpha )}{\sin (\theta )}}\right),$$ and additionally $\frac{\sin (\theta -\alpha )}{\sin (\theta )}+\frac{\sin (\alpha )}{\sin (\theta )} \geq 1$ for $0<\alpha<\theta<\pi$. But for general $a,b$, I am stuck.
Another observation is that the function $$g \colon \alpha \mapsto \frac{1}{\tanh \left(\frac{1}{\frac{\sin (\alpha )}{a \sin (\theta )}+\frac{\sin (\theta -\alpha )}{b \sin (\theta )}}\right)} - \bigg(\frac{\sin (\alpha )}{\tanh (a) \sin (\theta )}+\frac{\sin (\theta -\alpha )}{\tanh (b) \sin (\theta )}\bigg)$$ looks convex on $[0,\theta]$. If we could prove this, then we would be done because we know that $g(0) = g(\theta) = 0$. However, I am not sure how to prove this.
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*}