<s>[INST] in triangle ABC there satisfies equation: $\cos A \cos B+\sin A\sin B\sin C=1$, determine possible values of $C$ Question
in triangle ABC there satisfies equation: $\cos A\cos B+\sin A\sin B\sin C=1$, determine possible values of C
What I have so far
I've noticed that the given equation looked similar to $\cos(A-B)$ which was $\cos A\cos B+\sin A\sin B$. 
This I can extrapolate that $\cos(A-B)-\sin A\sin B+\sin A\sin B\sin C=1$
Thus $\cos(A-B)=1+\sin A\sin B+\sin A\sin B\sin C$
and when i factor out the common terms:
$\cos(A-B)=1+\sin A\sin B(1-\sin C)$
Right here I am not sure how to proceed and I got stuck
 [/INST] $\sin C=\frac{1-\cos A\cos B}{\sin A\sin B}\leq 1$
$\cos A\cos B+\sin A\sin B\geq 1$
$\cos\left(A-B\right)\geq 1$
$\cos\left(A-B\right)=1$
$A-B=0$
$A=B$
$\sin C=\frac{1-\cos^2 A}{\sin^2 A}=1$
$C=90$
$A=B=45$
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