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

I'd use the tangent half-angle substitution:
\begin{align*}
t&=\tan\frac\theta2 &
\sin\theta&=\frac{2t}{1+t^2} &
\tan\theta&=\frac{2t}{1-t^2} &
\sec\theta&=\frac{1+t^2}{1-t^2}
\\
u&=\tan\frac\phi2 &
\sin\phi&=\frac{2u}{1+u^2} &
\tan\phi&=\frac{2u}{1-u^2} &
\sec\phi&=\frac{1+u^2}{1-u^2}
\end{align*}
Then you have
\begin{align*}
a &= \frac{2 t^{2} u + 2 t u^{2} + 2 t + 2 u}{t^{2} u^{2} + t^{2} + u^{2} + 1} \\
b &= \frac{-2 t^{2} u - 2 t u^{2} + 2 t + 2 u}{t^{2} u^{2} -  t^{2} -  u^{2} + 1} \\
c &= \frac{-2 t^{2} u^{2} + 2}{t^{2} u^{2} -  t^{2} -  u^{2} + 1}
\end{align*}
\begin{multline*}
8bc=a[4b^2 + (b^2-c^2)^2] =\\
32\frac{t^{4} u^{3} + t^{3} u^{4} - t^{3} u^{2} - t^{2} u^{3} - t^{2} u - t u^{2} + t + u}{t^{4} u^{4} - 2 t^{4} u^{2} - 2 t^{2} u^{4} + t^{4} + 4 t^{2} u^{2} + u^{4} - 2 t^{2} - 2 u^{2} + 1}
\end{multline*}
The great benefit here is that after that first step, choosing to use that formulation, you almost don't have to think at all any more. From there on it's a straight-forward computation in rational functions. I left that to my computer algebra system, but if you are careful you can certainly do it by hand as well. The fact that my CAS expanded all those polynomials automatically makes the formulas above perhaps look more complicated than they really are, haven't tried.
