Finding value of determinant, Where $A+B+C+P+Q+R=0$ Finding value of Determinant $$\begin{vmatrix}
\tan (A+P) & \tan(B+P)  & \tan(C+P)\\\\ 
 \tan (A+Q)& \tan (B+Q)  & \tan (C+Q)\\\\ 
 \tan (A+R)&\tan (B+R)  & \tan(C+R) 
\end{vmatrix}$$ for all values of $A,B,C,P,Q,R$ Where $A+B+C+P+Q+R=0$
I did not understand how to start it, Could some help me, Thanks
 A: We have that if $X+Y+Z = 0$, then $\tan X+ \tan Y+\tan Z = \tan X \tan Y \tan Z$.
Hence for example $$\tan (A+P) \tan (B+Q) \tan (C+R) = \tan (A+P) +\tan (B+Q) +\tan (C+R)$$
Hence when you expand from R1, the determinant expands as 
$$\begin{align} & \left(\tan (B+Q) +\tan (C+R)-\tan (C+Q) - \tan (C+R)\right)\\
+ & \left(\tan (C+Q)+\tan (A+R) - \tan (A+Q)-\tan (C+R)\right)\\
+ & \left(\tan (A+Q)+\tan (B+R) - \tan (A+R) - \tan (B+Q)\right)\end{align} = 0$$
A: The determinant at hand evaluates to $0$ under given assumption $A+B+C+P+Q+R = 0$.
To prove this, we need an determinant identity. 
Let $[n] = \{ 1, 2, \ldots, n \}$. Given any $2n$ complex numbers $s_1,\ldots,s_n; t_1, \ldots, t_n$, construct two
$n \times n$ matrices $U$, $V$ whose entries at $j^{th}$ row and $k^{th}$ column have the form
$$U = \left(\frac{1}{s_j + t_k}\right)_{j,k\in [n]}
\quad\text{ and }\quad
V = \left(\frac{s_j - t_k}{s_j + t_k}\right)_{j,k\in [n]}$$
The determinant of $U$ is the so called Cauchy double alternant, one can show that (see this for a proof):
$$\det[U] = \frac{\prod_{1 \le j < k \le n}(s_j - s_k)(t_j - t_k)}{\prod_{j=1}^n\prod_{k=1}^n (s_j+t_k)}$$
As one can see below, what we really need is an expression of $\det[V]$. 
Padding $V$ to a $(n+1)\times(n+1)$ matrix and using linearity of determinant in  its first row, we have
$$\det[V] = 
\frac12\det \begin{bmatrix}
2 & 0_{1\times n}\\
-1_{n\times 1} & V
\end{bmatrix}
=
\frac12\det \begin{bmatrix}
1 & 1_{1\times n}\\
-1_{n\times 1} & V
\end{bmatrix}
+ 
\frac12\det \begin{bmatrix}
1 & -1_{1\times n}\\
-1_{n\times 1} & V
\end{bmatrix}
$$
For both matrices on RHS, adding first row to the remaining rows give us
$$\begin{align}
\det[V]
&=
\frac12\det \begin{bmatrix}
1 & 1_{1\times n}\\
0_{n\times 1} & \left(\frac{2s_j}{s_j + t_k}\right)_{j,k\in [n]}
\end{bmatrix}
+ 
\frac12\det \begin{bmatrix}
1 & -1_{1\times n}\\
0_{n\times 1} & \left(\frac{-2t_k}{s_j + t_k}\right)_{j,k\in [n]}
\end{bmatrix}\\
&= \frac12\det\left[\frac{2s_j}{s_j + t_k}\right]_{j,k\in [n]}
+ \frac12\det\left[\frac{-2t_k}{s_j + t_k}\right]_{j,k\in [n]}\\
&= 2^{n-1}\left(\prod_{j=1}^n s_j + (-1)^n \prod_{k=1}^n t_k\right)\det[U]
\end{align}
\tag{*1}
$$
In particular, for $n = 3$, this reduces to
$$\det\left[\frac{s_j - t_k}{s_j+t_k}\right]_{j,k\in [3]} = 4(s_1s_2s_3 - t_1t_2t_3)\det\left[\frac{1}{s_j+t_k}\right]_{j,k\in [3]}\tag{*2}$$
Back to original problem, if we change variables to $$(\theta_1,\theta_2,\theta_3, \phi_1, \phi_2,\phi_3) = (P,Q,R,-A,-B,-C)$$
The determinant at hand becomes
$$\begin{align}
\det\left[\tan(\theta_j - \phi_k)\right]_{j,k\in[3]}
&= \det
\left[\frac{\sin(\theta_j - \phi_k)}{\cos(\theta_j - \phi_k)}\right]_{j,k\in[3]}
= \det\left[-i
\frac{e^{i(\theta_j - \phi_k)} - e^{-i(\theta_j - \phi_k)}}
{e^{i(\theta_j - \phi_k)} + e^{-i(\theta_j - \phi_k)}}
\right]_{i,j\in[3]}\\
&= (-i)^3\det\left[
\frac{e^{2i\theta_j} - e^{2i\phi_k}}{e^{2i\theta_j} + e^{2i\phi_k}}
\right]_{j,k\in[3]}
\end{align}
$$
This has the form of $\det[V]$ with $s_j = e^{2i\theta_j}$ and $t_k = e^{2i\phi_k}$. By $(*2)$, this determinant contains the expression $s_1s_2s_3 - t_1t_2t_3$ as a factor.
Notice the condition $A + B + C + P + Q + R = 0$ implies
$$\sum_{j=1}^3 \theta_j = \sum_{k=1}^3 \phi_k
\quad\implies\quad
s_1s_2s_3 = e^{2i\sum_{j=1}^3\theta_j} = e^{2i\sum_{k=1}^3\phi_k} = t_1t_2t_3$$
As a result, the determinant at hand evaluates to $0$ under given assumption.
In general, for determinant of the form $\left[\tan(\alpha_j+\beta_k)\right]_{j,k\in [n]}$, we can start from $(*1)$ and simplify it trigonometrically to:
$$(-1)^{\lfloor\frac{n}{2}\rfloor}
\frac{\prod_{1\le j<k \le n}\sin(\alpha_j-\alpha_k)\sin(\beta_j-\beta_k)}{\prod_{j=1}^n\prod_{k=1}^n \cos(\alpha_j+\beta_k)}\times \begin{cases}
\cos\left(\sum_{\ell=1}^n(\alpha_\ell+\beta_\ell)\right), & n \text{ even}\\
\sin\left(\sum_{\ell=1}^n(\alpha_\ell+\beta_\ell)\right), & n \text{ odd}\\
\end{cases}
$$
Setting $(\alpha_1,\alpha_1,\alpha_3,\beta_1,\beta_2,\beta_3) = (P,Q,R,A,B,C)$, we find the determinant at hand equals to
$$- \frac{
\begin{array}{cl}
&\sin(A-B)\sin(A-C)\sin(B-C)\\
\times & \sin(P-Q)\sin(P-R)\sin(Q-R)
\end{array}
}{
\begin{array}{cl}
& \cos(A+P)\cos(A+Q)\cos(A+R)\\
\times & \cos(B+P)\cos(B+Q)\cos(B+R)\\
\times & \cos(C+P)\cos(C+Q)\cos(C+R)
\end{array}
}
\times \sin(A+B+C+P+Q+R)
$$
It is the last factor $\sin(A+B+C+P+Q+R)$ which forces the determinant at hand to vanish under the condition $A+B+C+P+Q+R = 0$.
A: We want to find the determinant of this Matrix.
$$
\begin{pmatrix}
Tan(A+P) & Tan(B+P) & Tan(C+P)\\
Tan(A+Q) & Tan(B+Q) & Tan(C+Q)\\
Tan(A+R) & Tan(B+R) & Tan(C+R)
\end{pmatrix}
$$
The determinant of this matrix gives us \to
$$
-Tan(A+R) \cdot Tan(B+Q) \cdot Tan(C+P)\\
+Tan(A+Q) \cdot Tan(B+R) \cdot Tan(C+P)\\
+Tan(A+R) \cdot Tan(B+P) \cdot Tan(C+Q)\\
-Tan(A+P) \cdot Tan(B+R) \cdot Tan(C+Q)\\
-Tan(A+Q) \cdot Tan(B+P) \cdot Tan(C+R)\\
+Tan(A+P) \cdot Tan(B+Q) \cdot Tan(C+R)
$$
But I still don't know how to depress it after this, even though you said
$
A+B+C+P+Q+R = 0
$
