Animated Solution for Mathematics - Trigonometry: Given $\alpha +\beta -\gamma =\pi$, prove that ${\sin }^2\alpha +{\sin }^2\beta -{\sin }^2\gamma =2\sin \alpha \sin \beta \cos \gamma$.

We are tasked with proving the trigonometric identity ${\sin }^2\alpha +{\sin }^2\beta -{\sin }^2\gamma =2\sin \alpha \sin \beta \cos \gamma$, given the constraint $\alpha +\beta -\gamma =\pi$.

This problem requires us to transform a sum of squared terms into a product. We will proceed by strategically grouping terms and utilizing the provided constraint to simplify the expression.

Our Left Hand Side (LHS) is ${\sin }^2\alpha +{\sin }^2\beta -{\sin }^2\gamma$. The presence of squared sine terms suggests the use of the identity:

We choose to group the last two terms: ${\sin }^2\beta -{\sin }^2\gamma$. Applying the identity, we obtain:

We must now incorporate the given condition $\alpha +\beta -\gamma =\pi$. Rearranging this constraint, we find:

Taking the sine of both sides, we apply the supplementary angle identity $\sin (\pi -\theta )=\sin \theta$:

This realization is the critical step that allows us to unify the variables in our expression.

Substituting $\sin \alpha$ for $\sin (\beta -\gamma )$ in our expression, the LHS becomes:

Observing the common factor of $\sin \alpha$, we factor it out:

To resolve the bracketed term, we substitute $\sin \alpha =\sin (\beta -\gamma )$ once more:

We now apply the sum-to-product identity $\sin (A-B)+\sin (A+B)=2\sin A\cos B$. Setting $A=\beta$ and $B=\gamma$, the bracket simplifies to $2\sin \beta \cos \gamma$.

Combining these results, we arrive at the final expression:

We have successfully reached the Right Hand Side. The identity is proved.