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
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
$$\begin{align} & \left(\tan (B+Q) +\tan (C+R)-\tan (C+Q) - \tan (C+R)\right)\\ + & \left(\tan (C+Q)+\tan (A+R) - \tan (A+Q)-\tan (C+R)\right)\\ + & \left(\tan (A+Q)+\tan (B+R) - \tan (A+R) - \tan (B+Q)\right)\end{align} = 0$$
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 $A+B+C+P+Q+R = 0$.
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 $
