\(\frac{\left(a\left(\sin c\right)^2-b\left(\sin c\right)\cos c\right)\left(a\left(\sin c\right)^2+b\left(\sin c\right)\cos c\right)-2a\left(\sin c\right)^2\left(\cos c\right)\left(d-a\cos c\right)}{\left(a\left(\sin c\right)^2-b\left(\sin c\right)\cos c\right)\left(a\left(\sin c\right)^2+b\left(\sin c\right)\cos c\right)\left(d-a\cos c\right)+2b^2a\left(\sin c\right)^4\cos c}=\frac{1}{d+a\cos c}\)

I am fairly sure these two expressions are equal, but I have no idea how to prove it. Any ideas?

Desmos somewhat confirms the equality here (if you scroll to the bottom of the input, the two values remain the same when you vary c).

If it helps, \(d=\sqrt{a^2-b^2}\).

## Comments

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TopNewestIs d equal to a^2-b^2 @Dan Ley – Neel Khare · 5 days, 1 hour ago

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– Dan Ley · 4 days, 22 hours 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 · 4 days, 22 hours 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 · 4 days, 22 hours 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 · 4 days, 22 hours ago

Thanks But I will tryLog in to reply

– Dan Ley · 4 days, 22 hours ago

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

– Neel Khare · 4 days, 22 hours 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 · 3 days, 15 hours ago

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@Ayush Rai @Rohit Camfar Thoughts? – Dan Ley · 5 days, 20 hours ago

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@Neel Khare – Ayush Rai · 5 days, 14 hours ago

I think the right person for this isLog in to reply