Question

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Please answer the above problem.

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Answer 1

$\large{\mathfrak{\underline{\underline{Answer:-}}}}$

Given:

$\sin( \alpha ) + \sin( \beta ) + \sin( \gamma ) - \sin( \alpha + \beta + \gamma )$

To Prove:

$= 4 \sin( \frac{ \alpha + \beta }{2} ) . \sin( \frac{ \beta + \gamma }{2} ). \sin( \frac{ \gamma + \alpha }{2} )$

$\large{\mathfrak{\underline{\underline{Explanation:-}}}}$

$L.H.S= ( \sin( \alpha ) + \sin( \beta ) - ( \sin( \alpha + \beta + \gamma ) - \sin \gamma )$

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

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

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

$= 2. \sin( \frac{ \alpha + \beta }{2} ) (2. \sin( \frac{ \alpha + \gamma }{2} ). \sin( \frac{ \beta + \gamma }{2} ))$

$= 4. \sin( \frac{ \alpha + \beta }{2} ) . \sin( \frac{ \beta + \gamma }{2} ) . \sin( \frac{ \gamma + \alpha }{2} )$

$\large{\mathfrak{\underline{\underline{Verification:-}}}}$

$\sin( \alpha ) + \sin( \beta ) + \sin( \gamma ) - \sin( \alpha + \beta + \gamma )$ $= 4 \sin( \frac{ \alpha + \beta }{2} ) . \sin( \frac{ \beta + \gamma }{2} ). \sin( \frac{ \gamma + \alpha }{2} )$

$\large{\boxed{\tt{Hence\;Proved!!}}}$

Answer 2

pls refer to the attachment pls!