Q. If $\alpha + \beta - \gamma = \pi$, then $\sin^{2}{\alpha + \sin^{2}{\beta - \sin^{2}\gamma}}$ is equal to
A) $2\sin{\alpha\sin{\beta\cos\gamma}}$
B) $2\cos\alpha\cos\beta\cos\gamma$
C) $2\sin\alpha\sin\beta\sin\gamma$
D) None of the above
Explanation
(a)
$\sin^{2}{\alpha{+ \sin^{2}}{\beta{- \sin^{2}}\gamma}}$
$= \sin^{2}{\alpha + \sin{(\beta - {\gamma)\sin{(\beta + \gamma)}}}}$
$= \sin^{2}{\alpha\sin{(\pi - \alpha)\sin{(\beta + \gamma)}\ \ \ \ \ \lbrack\because\ \alpha + \beta - \gamma = \pi\rbrack}}$
$= \sin{\alpha\lbrack\sin{\alpha + \sin{(\beta + \gamma)}}\rbrack}$
$= \sin{\alpha\lbrack\sin{\left\{ \pi - (\beta - \gamma) \right\} + \sin{(\beta + \gamma)}}\rbrack}$
$= \sin{\alpha\lbrack\sin{(\beta - \gamma) + \sin{(\beta + \gamma)}}\rbrack}$
$= \sin{\alpha\lbrack 2\sin{\beta\cos\gamma}\rbrack}$
$= 2\sin{\alpha\sin{\beta\cos\gamma}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.