cos2A.cos3A-cos2A.cos7A/sin4A.sin3A-sin2A.sin5A=sin7A+sin3A/sinA
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Step-by-step explanation:
(cos2Acos3A-cos2Acos7A+cosAcos10A)/(sin4Asin3A-sin2Asin5A+sin4Asin7A)
=(2cos2Acos3A-2cos2Acos7A+2cosAcos10A)/(2sin4Asin3A-2sin2Asin5A+2sin4Asin7A)
=(cos5A+cosA-cos9A-cos5A+cos11A+cos9A)/(cosA-cos7A-cos3A+cos7A+cos3A-cos11A)
=(cosA+cos11A)/(cosA-cos11A)
=2cos6Acos5A/2sin6Asin5A
=cot5Acot6A
Hence proved.
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