Question

sin8A+sin2A/cos8A+cos2A=tan5A

Answer 1

Answer:

2sin(2A+4A/2)cos(2A-4A/2) + 2sin(6A+8A/2)cos(6A-8A/2)/2cos(2A+4A/2)cos(2A-4A/2)+2cos(6A+8A/2)cos(6A-8A/2)

------using identity sinA + sinB=2sin(A+B/2)cos(A-B/2)

cosA+cosB= 2cosA+B/2 cos A-B/2

2 sin3Acos(-A)+2sin7Acos(-A) /2cis3Acis(-A) +2cos7Acos(-A)

2cos(-A) (sin3A+sin7A)/2cos(-A) (cos3A+cos7A)

sin3a+sin7a/cos3a+cos7a

2sin3a+7a/2 cos3a-7a/2/2cos3a+7a/2cos3a-7a/2

sin5a cos -4a/cos5a cos-4a

sin5a/cos5a=tan5a

Answer 2

Step-by-step explanation:

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