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Prove that $ \text{arctan}\left(\frac{a+d}{c}\right)=2\text{arctan}\left(\frac{a}{c}\right) $ if $a, d, c$ are positive reals satisfying $$ a^4+a^2c^2+a^2d^2+2a^3d = c^2d^2 $$

(credit: bobthesmartypants)

Martin Sleziak
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user1380792
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1 Answers1

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Try to arrive at the equation that needs to be proved from the given equation. $$ a^4+a^2c^2+a^2d^2+2a^3d = c^2d^2 $$ $$a^2(a^2+d^2+2ad)=c^2d^2-a^2c^2$$ $$(a^2)(a+d)^2+c^2(a^2-d^2)=0$$ $$(a+d)(a^2(a+d)+c^2(a-d))=0$$ $(a+d)$ cannot be equal to zero as $a,d$ are positive real numbers.So: $$a^2(a+d)+c^2(a-d)=0$$ $$\frac{-a^2}{c^2}=\frac{a-d}{a+d}$$ Adding $1$ to both sides: $$1-\frac{a^2}{c^2}=\frac{2a}{a+d}$$ Dividing both sides by $c$ and with a little rearrangement we get : $$\frac{a+d}{c}=\frac{\frac{2a}{c}}{1-\frac{a^2}{c^2}}$$ Now apply $ \text{arctan}$ on both the sides and use the formula $2\text{arctan}\left(x\right)=\text{arctan}\left(\frac{2x}{1-x^2}\right)$: $$ \text{arctan}\left(\frac{a+d}{c}\right)=2\text{arctan}\left(\frac{a}{c}\right) $$

lsp
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    Why don't you make your post all at once instead of continuously editing.. – user1380792 Feb 13 '14 at 19:22
  • So that it makes you think simultaneously along side. And if you get the solution mid way of me posting my answer, wouldn't it help us both ? – lsp Feb 13 '14 at 19:45