<s>[INST] find the value of given trigonometric equation If $\alpha$ and $\beta$ are two different values of $\theta$ which satisfy  $$bc \cos{​θ} \cos{\phi} + ac \sin{\theta} \sin{\phi} = ab,$$ then what is the value of $$(b^2 + c^2 - a^2) \cos{\alpha} \cos{\beta} + (c^2 + a^2 - b^2) \sin{\alpha} \sin{​\beta}\,?$$
 [/INST] Considering an ellipse
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
The equations of tangents at $[\alpha]$ and $[\beta]$ are
$$\frac{x\cos \alpha}{a}+\frac{y\sin \alpha}{b}=1$$
$$\frac{x\cos \beta}{a}+\frac{y\sin \beta}{b}=1$$
If the tangents meet at $(c\cos \phi, \, c\sin \phi)$, then
$$(c\cos \phi , \, c\sin \phi)=
\left(
  a\frac{\cos \frac{\alpha+\beta}{2}}{\cos \frac{\alpha-\beta}{2}} \, , \;
  b\frac{\sin \frac{\alpha+\beta}{2}}{\cos \frac{\alpha-\beta}{2}}
\right)$$
Refer to this link
I
\begin{align*}
  \frac{c^2\cos^2 \phi}{a^2}+\frac{c^2\sin^2 \phi}{b^2} &=
  \sec^2 \frac{\alpha-\beta}{2} \\
  \cos^2 \frac{\alpha-\beta}{2} &=
  \frac{a^2b^2}{c^2(a^2\sin^2 \phi+b^2\cos^2 \phi)} \\
  \cos (\alpha-\beta) &=
  \frac{2a^2b^2}{c^2(a^2\sin^2 \phi+b^2\cos^2 \phi)}-1 \\
\end{align*}
II
\begin{align*}
  \tan \frac{\alpha+\beta}{2} &=
  \frac{a}{b} \tan \phi \\
  \cos (\alpha+\beta) &=
  \frac{1-\frac{a^2}{b^2} \tan^2 \phi}{1+\frac{a^2}{b^2} \tan^2 \phi} \\
  &= \frac{b^2\cos^2 \phi-a^2\sin^2 \phi}{a^2\sin^2 \phi+b^2\cos^2 \phi}
\end{align*}
III
\begin{align*}
  E &=
  (b^2+c^2-a^2)\cos \alpha \cos \beta+
  (c^2+a^2-b^2)\sin \alpha \sin \beta \\[5pt]
  &= c^2\cos (\alpha-\beta)+(b^2-a^2)\cos (\alpha+\beta) \\[5pt]
  &= \frac{a^4\sin^2 \phi+a^2b^2+b^4\cos^2 \phi}
          {a^2\sin^2 \phi+b^2\cos^2 \phi}-c^2 \\[5pt]
  &= \frac{a^4\sin^2 \phi+
           a^2b^2\color{red}{(\sin^2 \phi+\cos^2 \phi)}+
           b^4\cos^2 \phi}
           {a^2\sin^2 \phi+b^2\cos^2 \phi}-c^2 \\[5pt]
  &= \fbox{$a^2+b^2-c^2$}
\end{align*}


Note Briefly:
  
  
*
  
*For real distinct $\alpha$ and $\beta$, $$c > \frac{ab}{\sqrt{a^2\sin^2 \phi+b^2\cos^2 \phi}} \ge \min(a,b) > 0$$
  
*If $c^2=a^2+b^2$, then $x^2+y^2=c^2$ is the director circle of the ellipse so that $AC \perp BC$.

</s>