Paso 1: Denotar las raíces:
$$ r_1=\tan\theta+\sen\theta,\qquad r_2=\tan\theta-\sen\theta. $$

Paso 2: Usar Vieta:
$$ r_1+r_2=-\frac{b}{a},\qquad r_1r_2=\frac{c}{a}. $$

Paso 3: Calcular suma y producto:
$$ r_1+r_2=2\tan\theta \Rightarrow 2\tan\theta=-\frac{b}{a} \Rightarrow \tan\theta=-\frac{b}{2a}. $$
$$ r_1r_2=(\tan\theta+\sen\theta)(\tan\theta-\sen\theta)=\tan^{2}\theta-\sen^{2}\theta. $$

Paso 4: Expresar $\sen^{2}\theta$ en función de $t=\tan\theta$:
$$ \sen^{2}\theta=\frac{\tan^{2}\theta}{1+\tan^{2}\theta}=\frac{t^{2}}{1+t^{2}}. $$
Entonces:
$$ r_1r_2=t^{2}-\frac{t^{2}}{1+t^{2}} =t^{2}\left(1-\frac{1}{1+t^{2}}\right) =t^{2}\left(\frac{t^{2}}{1+t^{2}}\right)=\frac{t^{4}}{1+t^{2}}. $$
Como $r_1r_2=\dfrac{c}{a}$:
$$ \frac{c}{a}=\frac{t^{4}}{1+t^{2}}. $$

Paso 5: Sustituir $t=-\dfrac{b}{2a}$ (solo aparece al cuadrado):
$$ t^{2}=\frac{b^{2}}{4a^{2}},\qquad t^{4}=\frac{b^{4}}{16a^{4}}. $$
Entonces:
$$ \frac{c}{a}=\frac{\frac{b^{4}}{16a^{4}}}{1+\frac{b^{2}}{4a^{2}}} =\frac{\frac{b^{4}}{16a^{4}}}{\frac{4a^{2}+b^{2}}{4a^{2}}} =\frac{b^{4}}{16a^{4}}\cdot\frac{4a^{2}}{4a^{2}+b^{2}} =\frac{b^{4}}{4a^{2}(4a^{2}+b^{2})}. $$
Multiplicando por $a$:
$$ c=\frac{b^{4}}{4a(4a^{2}+b^{2})}. $$
Reordenando:
$$ b^{4}=4ac(4a^{2}+b^{2}). $$

Resultado final: Queda demostrada la igualdad.