(cos7A+cos3A-cos5A-cosA)÷(sin7A-sin3A-sin5A+sinA)=cot2A
Answers
Answer:
{ ( 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
hey, here is your answer....
(cos7A + cos3A - cos5A - cosA)÷(sin7A - sin3A - sin5a + sinA)
By using transformations,
[(cos7A - cos5A) + (cos3A - cosA)]÷[(sin7A - sin5A) - (sin3A - sinA)]....
we get...
=[(2sin6A.sinA)+(2sin2A.sinA)]÷[(2cos6A.sinA)-(2cos2A.sinA)]...
=[{2sinA(sin6A+sin2A)}]÷[{2sinA(cos6A-cos2A)]
=[{2sin4A.cos2A}÷{2sin4A.sin2A}]
=[(cos2A)÷(sin2A)]
=cot2A....
Hence proved
hope this helps you...
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