<s>[INST] How to prove this trig. identity???(question 1 : solved)(question 2 : solved) question 1:
We know :
$$
\left\{
\begin{array}{ll}
\alpha \ + \beta\ +\theta\ = \pi\\ 
\alpha\ = \frac{\pi}{2} \\ 
\end{array} 
\right.
$$
How to prove this :
$$ 
\sin(\alpha)\sin(\beta)\sin(\alpha\ -\ \beta) + \sin(\beta)\sin(\theta)\sin(\beta\ -\ \theta) + \sin(\theta)\sin(\alpha)\sin(\theta\ -\ \alpha)+ \\
\sin(\alpha\ -\ \beta)\sin(\beta\ -\ \theta)\sin(\theta\ -\ \alpha) = 0 
$$

question 2:
And if we know that :
$$
\left\{
\begin{array}{ll}
\alpha \ + \beta\ +\theta\ = \pi\\ 
\cot(\alpha) \ + \cot(\beta)\ + \cot(\theta) \ = 2 \\ 
\end{array} 
\right.
$$
How to prove this:
$$1+\cos(\alpha)\cos(\beta)\cos(\theta) = 2\times \sin(\alpha)\sin(\beta)\sin(\theta)$$
Thanks a lot.
 [/INST] Question 2 :
The relation $\alpha + \beta + \theta = \pi$ implies that $\cos (\alpha + \beta + \theta) = -1$. But 
\begin{align}
\cos(\alpha + \beta + \theta) &= \cos\alpha \cos(\beta + \theta) - \sin \alpha\sin(\beta + \theta) \\
&= \cos(\alpha)(\cos\beta\cos\theta - \sin\beta \sin\theta) - \sin\alpha(\sin\beta \cos\theta + \sin\theta \cos\beta) \\ 
&= \cos\alpha \cos\beta \cos\theta - \cos\alpha\sin\beta\sin\theta - \sin\alpha\sin\beta\cos\theta - \sin\alpha\sin\theta\cos\beta,
\end{align}
so
$$1 +\cos\alpha\cos\beta\cos\theta = \cos\alpha\sin\beta\sin\theta + \sin\alpha\sin\beta\cos\theta + \sin\alpha\sin\theta\cos\beta.$$
Therefore
$$\frac{1}{\sin\alpha\sin\beta\sin\theta} + \cot\alpha\cot\beta\cot\theta = \cot\alpha + \cot\beta + \cot\theta = 2$$
and the result follows.
</s>