( cos 3a - cos a/sin 3a - sin a) + (cos 2a - cos 4a/ sin4a - sin 2a) = [sin a/ cos 2a cos 3a]

guys.. this problem was very difficult for me.. i almost tried everything.. pls help me.. mention the reason for each step.. pls help me.!!!!

... LHS

= [ ( cos 3a - cos a ) / ( sin 3a - sin a ) ] + [ ( cos 2a - cos 4a ) / ( sin 4a - sin 2a ) ]

= [ ( -2 sin 2a sin a ) / ( 2 cos 2a sin a ) ] + [ ( 2 sin 3a sin a ) / ( 2 cos 3a sin a ) ]

= ( - sin 2a / cos 2a ) + ( sin 3a / cos 3a )

= ( sin 3a / cos 3a ) - ( sin 2a / cos 2a )

= ( sin 3a cos 2a - cos 3a sin 2a ) / ( cos 2a cos 3a )

= [ sin ( 3a - 2a ) ] / ( cos 2a cos 3a )

= ( sin a ) / ( cos 2a cos 3a )

= RHS ........................................… Q.E.D.
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( sin 3a cos 2a - cos 3a sin 2a ) / ( cos 2a cos 3a )
how did that become
[ sin ( 3a - 2a ) ] / ( cos 2a cos 3a )
what happened to cos 2a - cos 3a??

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( sin 3a cos 2a - cos 3a sin 2a ) / ( cos 2a cos 3a )
how did that become
[ sin ( 3a - 2a ) ] / ( cos 2a cos 3a )
what happened to cos 2a - cos 3a??

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sin A cos B - cos A sin B = sin ( A - B )

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sin A cos B - cos A sin B = sin ( A - B )

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