How do I prove this seemingly simple trigonometric identity?

[Ganesh Ujwal]

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


I tried first method, like this:

\(\displaystyle a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2})\)
\(\displaystyle b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}\)
\(\displaystyle c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}\)
\(\displaystyle \\a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]}{\cos(\theta+\phi)+1}\\\cos(\theta-\phi)=\frac{ca}{b}-1\\\sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca+b)}\)

I tried second method, like this:

\begin{align*}4\cos^2(\theta)\cos^2(\phi)b^2 &= 4\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))^2 \\
&= 4(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))^2 \\
&= 4\sin^2(\theta + \phi) \\
&= 16\sin^2((\theta + \phi)/2)\cos^2((\theta + \phi)/2) \\
&= 16(1 - \cos^2((\theta + \phi)/2))\cos^2((\theta + \phi)/2) \\
&= 16\cos^2((\theta + \phi)/2) - 16\cos^4((\theta + \phi)/2)\end{align*}
\begin{align*}c^2 - b^2 &= \sec^2\theta + \sec^2\phi + 2\sec(\theta)\sec(\phi) - (\tan^2\theta + \tan^2\phi + 2\tan(\theta)\tan(\phi)) \\
&= 2 + 2\sec(\theta)\sec(\phi) - 2\tan(\theta)\tan(\phi)\end{align*}
\begin{align*}\cos(\theta)\cos(\phi)(c^2 - b^2) &= 2(\cos(\theta)\cos(\phi) + 1 - \sin(\theta)\sin(\phi)) \\
&= 2(1 + \cos(\theta + \phi)) \\
&= 4\cos^2((\theta + \phi)/2)\end{align*}
\begin{align*}cos^2(\theta)\cos^2(\phi)(b^2 - c^2)^2 = 16\cos^4((\theta + \phi)/2\end{align*}
\begin{align*}cos^2(\theta)\cos^2(\phi)(4b^2 + (b^2 - c^2)^2) = 16\cos^2((\theta + \phi)/2\end{align*}
\begin{align*}8\cos^2(\theta)\cos^2(\phi)bc/a &= 8\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))(\sec(\theta) + \sec(\phi))/(\sin(\theta) + \sin(\phi)) \\
&= 8(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))(\cos(\theta) + \cos(\phi))/(\sin(\theta) + \sin(\phi)) \\
&= 8\sin(\theta + \phi)(\cos(\theta) + \cos(\phi))/(\sin(\theta) + \sin(\phi)) \\
&= \frac{8(2\sin((\theta + \phi)/2)\cos((\theta + \phi)/2))(2\cos((\theta + \phi)/2)\cos((\theta - \phi)/2))}{2\sin((\theta + \phi)/2)\cos((\theta - \phi)/2)} \\
&= 16\cos^2((\theta+\phi)/2)\end{align*}