JEE Advance - Mathematics (1980 - No. 8)
Given $$\alpha + \beta - \gamma = \pi ,$$ prove that
$$\,{\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = 2\sin \alpha {\mkern 1mu} \sin \beta {\mkern 1mu} \cos y$$
$$\,{\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = 2\sin \alpha {\mkern 1mu} \sin \beta {\mkern 1mu} \cos y$$
This equation is incorrect; the correct form involves cosines, not sines.
This equation is a direct application of the Law of Sines in a triangle.
The provided equation is a trigonometric identity derived from the given condition and sine/cosine relationships.
This identity only holds true for specific values of alpha, beta, and gamma.
The equation is derived from complex number representation of trigonometric functions.