Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions1 Mark
Question
In $\triangle ABC$, prove that $( a - b ) \sin C +( b - c ) \sin A +( c - a ) \sin B =0$
✓
Answer
$\text { L.H.S. }=( a - b ) \sin C +( b - c ) \sin A +( c - a ) \sin B$
$=( a \sin C - b \sin C )+( b \sin A - c \sin A )+( c \sin B - a \sin B )$
$=( a \sin C - c \sin A )+( b \sin A - a \sin B )+( c \sin B - b \sin C )$
$=0+0+0=0=\text { R.H.S. }$
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