If $\sin(a)\sin(b)\sin(c)+\cos(a)\cos(b)=1$ then find the value of $\sin(c)$ If $$\sin(a)\sin(b)\sin(c)+\cos(a)\cos(b)=1,$$;where abc are the angles of the triangle.!
then find the value of $\sin(c)$. By trial and error put this triangle as right angled isosceles and got the answer..!! But i want a complete proof!
 A: Consider the two unit vectors on $S^2$ with spherical polar coordinates
$(a, 0)$ and $(b,\frac{\pi}{2}-c)$, its components in $\mathbb{R}^3$ are:
$$\begin{align}&(\sin a, 0, \cos a)\\
\text{ and }\quad&(\sin b \cos(\frac{\pi}{2}-c), \sin b\sin(\frac{\pi}{2}-c), \cos b)
=(\sin b\sin c,\sin b\cos c,\cos b)
\end{align}$$
The angle $\theta$ between them satisfies:
$$\cos\theta = \cos a\cos b + \sin a\sin b\sin c$$
Geometrically, $\cos\theta = 1$ if and only if these two unit vector coincides.
This in turn implies $a = b$ and $c = \frac{\pi}{2}$. By sine law, the two sides opposites to
angle $a$, $b$ have equal length. So the triangle is an right angled isosceles triangle and $\sin c = 1$.
Alternatively, one can use the identity
$$\begin{align} & (\sin a-\sin b \sin c)^2+(\sin b\cos c)^2 + (\cos a-\cos b)^2\\
= & 2 \left( 1 - ( \cos a\cos b + \sin a \sin b\sin c)\right)
\end{align}$$
to conclude $\cos c = 0$ and hence $\sin c = 1$.
A: $$\begin{align}
\sin(c)&=\frac{1-\cos(a)\cos(b)}{\sin(a)\sin(b)}\\
\sin(c)&=\frac{1}{\sin(a)\sin(b)}-\frac{\cos(a)\cos(b)}{\sin(a)\sin(b)}
\end{align}$$
Can you simplify from here?
A: Conventionally angles are written in uppercase, the sides are denoted in lower case
As $0<A,B,C< \pi,0 <\sin A,\sin B,\sin C\le1$
$$\implies \sin A\sin B\sin C+\cos A\cos B\le \sin A\sin B+\cos A\cos B=\cos(A-B)  $$
$$\implies 1\le \cos(A-B)$$ which is possible iff $\cos(A-B)=1\iff A-B=2n\pi$ where $n$ is any integer 
But as $0<A,B< \pi, -\pi<A-B<\pi\implies A-B=0\iff A=B$
Then from the given relation, we have $\displaystyle \sin^2A\sin C+\cos^2A=1\iff \sin^2A\sin C=1-\cos^2A=\sin^2A$
$\displaystyle\implies \sin C=1$ as $\sin A\ne0$ as $0<A<\pi$
