Question

If $\alpha +\beta -\gamma$ = $\pi$, then $si{n}^2\alpha +si{n}^2\beta -si{n}^2\gamma$ is equal to

• A. $2\cos \alpha \cos \beta \cos \gamma$.
• B. $2\sin \alpha \sin \beta \sin \gamma$
• C. $2\sin \alpha \sin \beta \cos \gamma$.
• D. $2\cos \alpha \cos \beta \sin \gamma$

Answer 1

The correct option is C

$2\sin \alpha \sin \beta \cos \gamma$.

It is given that $\alpha$ + $\beta$ - $\gamma$ = $\pi$
To find : $si{n}^2\alpha$ + $si{n}^2\beta$ - $si{n}^2\gamma$

As, we have done earlier problems we should reduce second power of sine function to power 1 using formula

cos2A = 1 - 2 $si{n}^{2}A$

or

$si{n}^2\beta$ - $si{n}^2\gamma$ can be expanded using identity $a^2$ - $b^2$ = (a + b)(a - b)

$si{n}^2\alpha$ + $si{n}^2\beta$ - $si{n}^2\gamma$

= $si{n}^2\alpha$ + ($sin\beta$ + $sin\gamma$) ($sin\beta$ - $sin\gamma$)

= $si{n}^2\alpha +\{2sin\frac{(\beta +\gamma )}{2}.cos\frac{(\beta -\gamma )}{2}\}\{2cos\frac{(\beta +\gamma )}{2}.sin\frac{(\beta -\gamma )}{2}\}$

= $si{n}^2\alpha +\{2sin\frac{(\beta +\gamma )}{2}.cos\frac{(\beta +\gamma )}{2}.2sin\frac{(\beta -\gamma )}{2}.cos\frac{(\beta -\gamma )}{2}\}$

= $si{n}^2\alpha +\{sin2\frac{(\beta +\gamma )}{2}.sin2\frac{(\beta -\gamma )}{2}\}$ {sin 2A = 2 sinA cosA}

= $si{n}^2\alpha$ + sin $(\beta +\gamma )$.sin $(\beta -\gamma )$
$\{\beta -\gamma =\pi -\alpha \}$

= $si{n}^2\alpha$ + sin $(\beta +\gamma )$.sin $(\pi -\alpha )$

= $si{n}^2\alpha$ + sin $(\beta +\gamma )$ sin $\alpha$

= sin $\alpha$ [sin $\alpha$ + sin $(\beta +\gamma )$]

= sin $\alpha$ $[2sin\frac{(\alpha +\beta +\gamma )}{2}.cos\frac{(\alpha -\beta -\gamma )}{2}]$
$\because \left\{\begin{array}{cc}\alpha +\beta =\pi +\gamma & \\ \alpha -\gamma =\pi -\beta & \end{array}\right\}$
= sin $\alpha$ $[2sin\frac{(\pi +\gamma +\gamma )}{2}.cos\frac{(\pi -\beta -\beta )}{2}]$

= $sin\alpha [2sin(\frac{\pi }{2}+\gamma ).cos(\frac{\pi }{2}+\beta )]$

= sin $\alpha$ [2 cos$\gamma$.sin$\beta$]

= 2 sin $\alpha$ sin $\beta$ cos$\gamma$.
Hence the correct answer is Option A.