Question

$prove \: that \: \frac{sin8a \: cosa - sin6a \: cos3a}{cos2a \: cosa - sin4a \: sin3a} = tan2a$


⭐No copied answers please
⭐Step-by-step explanation needed.. ​

Answer 1

Step-by-step explanation:

LHS=(sin8a. cosa-sin6a.cos3a)/(cos2a.cosa-sin4a.sin3a)

=(2sin8a.cosa-2sin6a.cos3a)/(2cos2a.cosa-2sin4a.sin3a)

=(sin9a+sin7a-sin9a-sin3a)/(cos3a+cosa-cosa+cos7a)

=(sin7a-sin3a)/(cos3a+cos7a)

=2cos5asin2a/2cos5acos2a

=tan2a=RHS

LHS=RHS(Proved)

Answer 2

Answer:

I tried my best‼️‼️‼️

Refer the attachement for better understanding

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