Question

Cos2A cos3A - cos2Acos7A +cosAcos10A\sin4Asin3A-sin2Asin5A +sin4Asin7A= cot6Acot5A

Answer 1

Answer:

Step-by-step explanation:

(cos2Acos3A-cos2Acos7A+cosAcos10A)/(sin4Asin3A-sin2Asin5A+sin4Asin7A)

//multiply and divide by 2 both numerator and denominator

=(2cos2Acos3A-2cos2Acos7A+2cosAcos10A)/(2sin4Asin3A-2sin2Asin5A+2sin4Asin7A)

//2cosAcosB = Cos(A+B) + Cos(A-B)

//2SinASinB = Cos(A-B) - Cos(A+B)

=(cos5A+cosA-cos9A-cos5A+cos11A+cos9A)/(cosA-cos7A-cos3A+cos7A+cos3A-cos11A)

=(cosA+cos11A)/(cosA-cos11A)

=2cos6Acos5A/2sin6Asin5A

=cot5Acot6A

= R.H.S

Hence Proved