Math, asked by soorajseenha8, 3 months ago

cos5A/sinA+sin5A/cosA=2cos4A.sin2A​

Answers

Answered by pratibhakumariprusty
1

Answer:

L.H.S.

= (cos5a.sin2a-cos4a.sin3a)/(sin5a.sin2a-cos4a.cos3a)

Multiply numerator and denominator by 2.

= 2(cos5a.sin2a - cos4a.sin3a) / 2(sin5a.sin2a - cos4a.cos3a)

= (2cos5a.sin2a - 2cos4a.sin3a) / (2sin5a.sin2a - 2cos4a.cos3a)

= [sin(5a+2a)-sin(5a-2a)-sin(4a+3a)+sin(4a-3a)]/[cos(5a-2a)-cos(5a+2a)-sin(4a-3a)+cos(4a+3a)]

= (sina - sin3a)/(cso3a-cosa)

= (- 2cos2a.sina)/(-2sin2a.sina)

= cos2a/sin2a

= cot2a

= R.H.S.

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