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To prove that Cos 2B /b^2 - Cos 2A/ a^2 = a^2 - b^2 / a^2 b^2

To prove that Cos 2B /b^2 - Cos 2A/ a^2 = a^2 - b^2 / a^2 b^2

 

Grade:11

1 Answers

Arun
25763 Points
3 years ago
Hello 

First use the formula cos 2A = 1 - 2sin²A 

(1 - 2sin²A)/a² - (1 - 2sin²B)/b² = 1/a² - 2sin²A/a² - 1/b² + 2sin²B/b² 

= 1/a² - 1/b² + 2sin²B/b² - 2sin²A/a² 

Now think of the law of sines in a triangle 

sinA/a = sinB/b 
sin²A/a² = sin²B/b² 
Hence 
2sin²B/b² - 2sin²A/a² = 0 

So 

Cos2A/a² - cos2B/b² = 1/a² - 1/b² 
 
Hence 
 
Cos2B/a² - cos2A/b² = 1/b² - 1/a² = (a² - b²)/a² b²
 
Hence proved
 
Regards
Arun (askIITians forum expert)

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