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(Prove that):
(cos 7A + cos 3A - Cos 5A - COS A) / (sin 7A - sin 5A - sin3A + sinA) = cot 2A​

User Skiwi
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2 Answers

6 votes

Explanation:

this photo answer ...

i hope this will help

(Prove that): (cos 7A + cos 3A - Cos 5A - COS A) / (sin 7A - sin 5A - sin3A + sinA-example-1
User Patrik
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4.7k points
2 votes

Answer:

Explanation:

(cos7A + cos3A - cos5A - cosA)÷(sin7A - sin3A - sin5a + sinA)

By using transformations,

[(cos7A - cos5A) + (cos3A - cosA)]÷[(sin7A - sin5A) - (sin3A - sinA)]....

we get...

=[(2sin6A.sinA)+(2sin2A.sinA)]÷[(2cos6A.sinA)-(2cos2A.sinA)]...

=[{2sinA(sin6A+sin2A)}]÷[{2sinA(cos6A-cos2A)]

=[{2sin4A.cos2A}÷{2sin4A.sin2A}]

=[(cos2A)÷(sin2A)]

=cot2A

User Steven McConnon
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4.9k points