1.

If A+B+C=pi, then prove the following. sin2A+sin2B-sin2C=4cosA cdot cosB cdot sinC

Answer»

Solution :L.H.S. =sin2A+sin2B-SIN2C
`2sinfrac{2A+2B}{2}cosfrac{2A-2B}{2}-sin2C`
`=2SIN(A+B)COS(A-B)-2sinCcosC`
`=2sinCcos(A-B)-2sinCcosC`
`[becauseA+B=pi-C`
or, sin(A+B)=sin(pi-C)=sinC]`
`=2sinC[cos(A-B)-cosC]`
`=2sinC[cos(A-B)-cos(A+B)]`
`=2sinC cdot 2cosA cdot cosB`
`=4cosA cosBsinC=R.H.S`


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