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Question

Prove that sin A - sin 3A + sin 5A - sin 7A/ sin A - sin 3A - sin 5A +sin 7A = cot 2A.

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Solution

= (sin A - sin 3A + sin 5A - sin 7A)/( sin A - sin 3A - sin 5A +sin 7A)

= { ( sin A - sin 7A ) + ( sin 5A -sin 3A ) } / { ( cos A + cos 7A ) - ( cos 5A + cos 3A ) }

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

( sin A - sin B = 2 sin (A + B)^2 cos (A - B)^2)

(cos A - cos B = -2 sin (A + B)^2 sin (A - B)^2)

{cos (- theta )= cos theta and sin (- theta )= - sin theta}

= {2 cos 4A [ sin A - sin 3A ] } / { 2 cos 4A [ cos 3A - cos A]}

= { sin A - sin 3A } / { cos 3A - cos A }

= [ -2 sin A cos 2A ] / [ -2 sin 2A sin A ] = cos 2A / sin 2A = cot 2A = RHS

Hope u understood!!


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