I think you can do it by expanding the bottom, and multiplying the top out by \(d+acosc\), and showing that these two expressions are equal. But I can't see a better, nice method.

@Neel Khare don't worry about it, I just multiplied the equation out so that there were no fractions, and replaced \(\sin^2c\) with \(1-\cos^2c\). Then you can show that the two expressions are equal, but it's not fun or elegant or particularly difficult...

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TopNewestIs d equal to a^2-b^2 @Dan Ley

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I think you can do it by expanding the bottom, and multiplying the top out by \(d+acosc\), and showing that these two expressions are equal. But I can't see a better, nice method.

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I agree let me work on it I will get back to you after 2-3 days I am a little busy Board exams are coming up

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No, \(d^2=a^2-b^2\), so \(d=\sqrt{a^2-b^2}\).

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Ok

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@Neel Khare don't worry about it, I just multiplied the equation out so that there were no fractions, and replaced \(\sin^2c\) with \(1-\cos^2c\). Then you can show that the two expressions are equal, but it's not fun or elegant or particularly difficult...

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I think the right person for this is @Neel Khare

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