Question

cosa+cos3A+cos5A+cos7A/sinA+sin3A+sin5A+sin7A

Answer 1

Answer:

Cos7A + os3A = 2cos5Acos2A 

cos5A + cosA = 2cos3Acos2A 

(cos7A + os3A) - (cos5A + cosA) = 2cos2A(cos5A - cos3A) 

sin7A - sin3A = 2cos5Asin2A 

sin5A - sinA = 2cos3Asin2A 

(sin7A - sin3A) - (sin5A - sinA) = 2sin2A(cos5A - cos3A) 

(cos7A + os3A - cos5A - cosA)/(sin7A - sin3A - sin5A + sinA)

= {2cos2A(cos5A - cos3A)}/{2sin2A(cos5A - cos3A)} 

= cot2A

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Answer 2

Answer:

{(cosA+cos7A) + (cos3A+cos5A)}/{(sinA+sin 7A)+(sin3A+sin5A)}

(2cos4A.cos3A+2cos4A.cosA)/(2sin4A.cos3A+2sin4A.cosA)

2cos4A(cos3A+cosA)/{2sin4A(cos3A+cosA)}

=cot4A ans