Math, asked by SteveUday49, 1 year ago

CosA-cos3A-cos5A+cos7A/SinA-sin3A+sin5A-sin7A=Tan2A

Answers

Answered by bluelinebus4

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

SteveUday49: prove that Tan2A

SteveUday49: not cot 2 A

SteveUday49: ok thanks for that and I did that problem

Answered by suhaa6

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

suhaa6: 9..

suhaa6: hmm

suhaa6: ok

Previous Question

Next Question