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

Question

Arun · Answer

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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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

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