Math, asked by venkateswarammav3, 6 days ago

prove that sin6A+sin2A+ sin8A+sin4A /cos6A+cos2A+ cos8A+ cos4A=tan5A​

Answers

Answered by Anonymous
1

Answer:

sin5a/cos5a=tan5a

Step-by-step explanation:

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

#BrainLock .

Similar questions