Question

prove that sin7A+sinA÷cos5A-cos3A=sin2A-cos2A×cotA​

Answer 1

{ ( 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 - B2))

{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!!

please mark as the brainliest answer.

Answer 2

hey mate here is ur answer .

i hope it helps u