@Neel Khare
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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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Dan Ley
·
2 months, 1 week ago

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@Dan Ley
–
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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Neel Khare
·
2 months, 1 week ago

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@Neel Khare
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That's fine don't stress over it, it's just part of my school project:) Good luck with the exams bro.
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Dan Ley
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2 months, 1 week ago

@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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Dan Ley
·
2 months, 1 week ago

## Comments

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TopNewestIs d equal to a^2-b^2 @Dan Ley – Neel Khare · 2 months, 1 week ago

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– Dan Ley · 2 months, 1 week ago

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.Log in to reply

– Neel Khare · 2 months, 1 week ago

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 upLog in to reply

– Dan Ley · 2 months, 1 week ago

That's fine don't stress over it, it's just part of my school project:) Good luck with the exams bro.Log in to reply

– Neel Khare · 2 months, 1 week ago

Thanks But I will tryLog in to reply

– Dan Ley · 2 months, 1 week ago

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

– Neel Khare · 2 months, 1 week ago

OkLog in to reply

@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... – Dan Ley · 2 months, 1 week ago

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@Neel Khare – Ayush Rai · 2 months, 1 week ago

I think the right person for this isLog in to reply