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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 · 5 days, 1 hour 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. · 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 up · 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. · 4 days, 22 hours ago

Thanks But I will try · 4 days, 22 hours ago

No, $$d^2=a^2-b^2$$, so $$d=\sqrt{a^2-b^2}$$. · 4 days, 22 hours ago

Ok · 4 days, 22 hours 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... · 3 days, 15 hours ago

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

I think the right person for this is @Neel Khare · 5 days, 14 hours ago

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