Nice Arctan Identity 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)

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 $2\text{arctan}\left(x\right)=\text{arctan}\left(\frac{2x}{1-x^2}\right)$:
$$ \text{arctan}\left(\frac{a+d}{c}\right)=2\text{arctan}\left(\frac{a}{c}\right) $$
