I am trying to get a derivation of the spherical law of cosines. The Wikipedia page [https://en.wikipedia.org/wiki/Spherical_law_of_cosines ] contains a proof that I don't understand because there are not enough intermediate steps shown.
The Wikipedia page says that for unit vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ : $$ \begin{align} \cos(a) = \mathbf{u}\cdot\mathbf{v} \end{align} $$ \begin{align} \cos(b) = \mathbf{u}\cdot\mathbf{w} \end{align} $$ \begin{align} \cos(c) = \mathbf{v}\cdot\mathbf{w} \end{align} $$
The unit vector $\mathbf{t}_a$ is defined as the unit vector perpendicular to $\mathbf{u}$ in the u-v plane, whose direction is given by the component of v perpendicular to u. Wikipedia explains that $$ \begin{align} \mathbf{t}_a = { {\mathbf{v}-\mathbf{u}(\mathbf{u \cdot v})} \over { \left\lVert {\mathbf{v}-\mathbf{u}(\mathbf{u \cdot v})} \right\rVert }} = {{{\mathbf{v}-\mathbf{u}\cos(a)}} \over {\sin(a)}} \end{align} $$
Similarly, $$ \begin{align} {\mathbf{t}_b} = {{{\mathbf{w}-\mathbf{u}\cos(b)}} \over {\sin(b)}} \end{align} $$
Without any further justification, their proof ends by the claim: $$ \begin{align} {\mathbf{t}_a} \cdot {\mathbf{t}_b} = {{\cos(c)-\cos(a)\cos(b)} \over {\sin(a)\sin(b)}} \end{align} $$ It is the final step, which yeilds the formula for $\mathbf{t}_a\cdot\mathbf{t}_b$, that I do not comprehend. Can somebody please explain to me the justification for this expression?
$$\mathbf{t}_a \cdot \mathbf{t}_b = \frac{\mathbf{v} - \mathbf{u} \cos(a)}{\sin(a)} \cdot \frac{\mathbf{w} - \mathbf{u} \cos(b)}{\sin(b)} \\ = \frac{\mathbf{v} \cdot \mathbf{w} - \mathbf{v} \cdot \mathbf{u} \cos(b) - \mathbf{u} \cdot \mathbf{w} \cos(a) + \mathbf{u} \cdot \mathbf{u}\cos(a)\cos(b)}{\sin(a)\sin(b)} \\ = \frac{\cos(c) - \cos(a)\cos(b) - \cos(a)\cos(b) + \cos(a)\cos(b)}{\sin(a)\sin(b)} = \frac{\cos(c) - \cos(a)\cos(b)}{\sin(a)\sin(b)}.$$
Note that $\mathbf{u} \cdot \mathbf{u} = 1$ since it's a unit vector.