How do I prove this seemingly simple trigonometric identity

$$a = \sin\theta+\sin\phi\\b=\tan\theta+\tan\phi\\c=\sec\theta+\sec\phi$$ Show that, $8bc=a[4b^2 + (b^2-c^2)^2]$

I tried to solve this for hours and have gotten no-where. Here's what I've got so far : $$ \\a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2}) \\ b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)} \\c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)} \\a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]}{\cos(\theta+\phi)+1}\\\cos(\theta-\phi)=\frac{ca}{b}-1\\\sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca+b)}$$

Hint: Divide both sides of $8bc=a[4b^2 + (b^2-c^2)^2]$ by $bc$. You will end up with this: $$8 = a[4\frac{b}{c}+bc(\frac{b}{c}-\frac{c}{b})^2]$$ Take the RHS and you can prove that it is equal to 8.

Process: Calculate $\frac{b}{c}$ from the given equations. $$b = \frac{\sin (\theta + \phi)}{\cos \theta. \cos \phi}$$ $$c = \frac{\cos \theta + \cos \phi}{\cos \theta. \cos \phi}$$ $$\frac{b}{c} = \frac{2\sin(\frac{\theta + \phi}{2}).\cos(\frac{\theta + \phi}{2})}{2\cos(\frac{\theta + \phi}{2}).\cos(\frac{\theta - \phi}{2})}$$ $$\frac{b}{c} = \frac{\sin(\frac{\theta + \phi}{2})}{\cos(\frac{\theta - \phi}{2})}$$ You will end up with: $$\frac{b}{c}-\frac{c}{b}=-\frac{\cos \theta. \cos \phi}{\cos(\frac{\theta - \phi}{2}).sin(\frac{\theta + \phi}{2})}$$ $$bc(\frac{b}{c}-\frac{c}{b})^2 = \frac{\sin (\theta + \phi)(\cos \theta + \cos \phi)}{\cos^2 \theta. \cos^2 \phi}\frac{\cos^2 \theta. \cos^2 \phi}{\cos^2(\frac{\theta - \phi}{2}).\sin^2(\frac{\theta + \phi}{2})}$$ $$ = \frac{2\sin(\frac{\theta + \phi}{2}).\cos(\frac{\theta + \phi}{2})2\cos(\frac{\theta + \phi}{2}).\cos(\frac{\theta - \phi}{2})}{\cos^2(\frac{\theta - \phi}{2}).\sin^2(\frac{\theta + \phi}{2})}$$ $$ = \frac{4\cos^2 (\frac{\theta + \phi}{2})}{\cos(\frac{\theta - \phi}{2})\sin (\frac{\theta + \phi}{2})}$$ $$4\frac{b}{c} + bc(\frac{b}{c}-\frac{c}{b})^2 = \frac{4}{\cos (\frac{\theta - \phi}{2})}(\sin(\frac{\theta + \phi}{2})+\frac{\cos^2 (\frac{\theta + \phi}{2})}{\sin(\frac{\theta + \phi}{2})})$$ $$= \frac{4}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$ $$a[4\frac{b}{c} + bc(\frac{b}{c}-\frac{c}{b})^2] = 4\frac{\sin \theta + \sin \phi}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$ $$ = 4\frac{2\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$ $$= 8$$

$\color{red}{c^2=\sec^2\theta+\sec^2\phi+2\sec\theta\sec\phi\tag{1}}$

$\color{blue}{b^2=\tan^2\theta+\tan^2\phi+2\tan\theta\tan\phi\tag{2}}$

$(1)-(2)$ gives,

$\begin{align}\left(c^2-b^2\right) & =\sec^2\theta+\sec^2\phi+2\sec\theta\sec\phi-\tan^2\theta-\tan^2\phi-2\tan\theta\tan\phi\\ &=2(1+\sec\theta\sec\phi-\tan\theta\tan\phi)\\&=2\left(1+\dfrac{1}{\cos\theta\cos\phi}-\dfrac{\sin\theta\sin\phi}{\cos\theta\cos\phi}\right)\\&=\dfrac{2}{\cos\theta\cos\phi}(1+\cos(\theta+\phi))\\&=\dfrac{4\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos\theta\cos\phi}\end{align}$

$\color{darkgreen}{\therefore\left(c^2-b^2\right)^2=\dfrac{16\cos^4\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2\phi}\tag{3}}$

$\color{brown}{4b^2=\dfrac{4\sin^2(\theta+\phi)}{\cos^2\theta\cos^2 \phi}=\dfrac{16\sin^2\left(\dfrac{\theta+\phi}{2}\right)\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2 \phi}\tag{4}}$

$$\boxed{4b^2+\left(b^2-c^2\right)^2=\dfrac{16\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2 \phi}}$$

$c=\dfrac{\cos\theta+\cos\phi}{\cos\theta\cos\phi}=\dfrac{2\cos\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)}{\cos\theta\cos\phi}$

$b=\dfrac{\sin(\theta+\phi)}{\cos\theta\cos\phi}=\dfrac{2\sin\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta+\phi}{2}\right)}{\cos\theta\cos\phi}$

$a=2\sin\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)$

$$\boxed{\dfrac{8bc}{a}=\dfrac{32\sin\left(\dfrac{\theta+\phi}{2}\right)\cos^2\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)}{2\sin\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)\cos^2\theta\cos^2 \phi}=\dfrac{16\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2 \phi}}$$