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Trigonometry Help

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

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

Note by Dan Ley
5 days, 20 hours ago

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

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@Neel Khare 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. Dan Ley · 4 days, 22 hours 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 Neel Khare · 4 days, 22 hours ago

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@Neel Khare That's fine don't stress over it, it's just part of my school project:) Good luck with the exams bro. Dan Ley · 4 days, 22 hours ago

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@Dan Ley Thanks But I will try Neel Khare · 4 days, 22 hours ago

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@Neel Khare No, \(d^2=a^2-b^2\), so \(d=\sqrt{a^2-b^2}\). Dan Ley · 4 days, 22 hours ago

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@Dan Ley Ok Neel Khare · 4 days, 22 hours ago

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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... 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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@Dan Ley I think the right person for this is @Neel Khare Ayush Rai · 5 days, 14 hours ago

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