How prove this $\frac{1}{\cos{A}}-\frac{\sin{\frac{A}{2}}}{\sin{\frac{B}{2}}\sin{\frac{C}{2}}}=4$ in $\Delta ABC $ not An equilateral triangle, let
$$\begin{vmatrix}
2\sin{A}\sin{C}-\cos{B}&2\sin{A}\sin{B}-\cos{C}\\
4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+\cos{B}&4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+\cos{C}
\end{vmatrix}=0$$
prove that
$$\dfrac{1}{\cos{A}}-\dfrac{\sin{\dfrac{A}{2}}}{\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}}=4$$
my idea:
use this well known
$$4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+1=\cos{A}+\cos{B}+\cos{C}$$
then we have
$$\Longrightarrow \begin{vmatrix}
2\sin{A}\sin{C}-\cos{B}&2\sin{A}\sin{B}-\cos{C}\\
\cos{A}+\cos{C}+2\cos{B}-1&\cos{A}+\cos{B}+2\cos{C}-1
\end{vmatrix}=0$$
then
$$(2\sin{A}\sin{C}-\cos{B})(\cos{A}+\cos{B}+2\cos{C}-1)-(2\sin{A}\sin{B}-\cos{C})(\cos{A}+\cos{C}+2\cos{B}-1)=0$$
folowing  I can't work,Thank you everyone  can help
 A: I'm getting the result without using the formula. 
Note that, since $\Delta ABC$ is a triangle, $$2\sin A\sin C-\cos B=\cos (A-C)-\cos(A+C)-\cos C=\cos(A-C)$$
Similarly, $$2\sin A\sin B-\cos C=\cos (A-B)$$
Now, observe that if $B=C$, the determinant condition is automatically satisfied and in that case $$\large \cos A=-\cos 2B,\ \sin \frac{A}{2}=\cos B$$ So, if the answer holds then we will be able to find a solution for $B$ and hence for $C,A$! So the result is not a general one if $B=C$. So I'm assuming $B\ne C$. 
Now, from the determinant, and with the assumption stated above along with the above equalities, we get,\begin{equation}
\begin{split}
\ &4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\left(\cos(A-C)-\cos(A-B)\right)\\
\ =&\cos B\cos(A-B)-\cos C \cos(A-C)\\
\  \Rightarrow & 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}2\sin\left(\frac{C-B}{2}\right)\sin\left(\frac{3A-\pi}{2}\right)\\
\ =&\frac{1}{2}\left(\cos A+\cos(A-2B)-\cos A-\cos (A-2C)\right)
\ =&\sin(C-B)\sin 2A\\
\ \Rightarrow & 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\cos\left(\frac{3A}{2}\right)\\
\ =&-\cos\left(\frac{B-C}{2}\right)\sin 2A\\
\ \Rightarrow & \sin \frac{B}{2}\sin \frac{C}{2}\cos\left(\frac{3A}{2}\right)\\
\ =& -\cos\left(\frac{B-C}{2}\right)\cos\left(\frac{A}{2}\right)\cos A\\
\end{split}
\end{equation}
So, now we get \begin{equation}
\begin{split}
\large
\frac{\cos \frac{3A}{2}}{\cos \frac{A}{2} \cos A}=& -\frac{\cos \frac{B-C}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}\\
\Rightarrow \frac{\cos \frac{3A}{2}-\cos \frac{A}{2} \cos A}{\cos \frac{A}{2} \cos A}=& \frac{-\cos \frac{B-C}{2}-\sin \frac{B}{2}\sin \frac{C}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}\\
\Rightarrow -\frac{\sin A \sin \frac{A}{2}}{\cos \frac{A}{2} \cos A}=&\frac{-\cos \frac{B}{2}\cos \frac{C}{2}-2\sin \frac{B}{2}\sin \frac{C}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}\ \\
\Rightarrow \frac{2\sin^2 \frac{A}{2}}{\cos A}= &\frac{\cos \frac{B}{2}\cos \frac{C}{2}}{\sin \frac{B}{2} \sin \frac{C}{2}}+2\\
\Rightarrow \frac{1}{\cos A}=3+\cot \frac{B}{2} \cot \frac{C}{2}
\end{split}
\end{equation}
Now, observe that $$\sin \frac{A}{2}=\cos \frac{B+C}{2}=\cos \frac{B}{2}\cos \frac{C}{2}-\sin \frac{B}{2}\sin \frac{C}{2}\\
\Rightarrow \frac{\sin \frac{A}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}=\cot \frac{B}{2} \cot \frac{C}{2}-1$$
 Hence we get $$\Rightarrow \frac{1}{\cos A}=4+\frac{\sin \frac{A}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}$$
A: Hint:
Your identity is equivalent to:
$$\frac{\sin(A/2)\sin(B/2)\sin(C/2)}{\cos A} + \sin^2(A/2) = 4\sin(A/2)\sin(B/2)\sin(C/2)$$
$$\frac{1}{2}\left(1-\cos(A)\right)=\left(\cos(A)+\cos(B)+\cos(C)-1\right)\left(1 - \frac{1}{\cos(A)} \right)$$
Notice that this form is quite close to the fraction you get from the determinant. 
