Let us have..... D = %26%20tan(B%2BP)%20%20%26tan(C%2BP)%20\\%20%20tan(A%2BQ)%26%20%20tan(B%2BQ)%20%26%20%20tan(C%2BQ)\\%20%20%20tan(A%2BR)%26%20tan(B%2BR)%20%20%26%20tan(C%2BR)%20\end{vmatrix})
Now as it is given,,,,
A+B+C+P+Q+R = 0 rewriting it as (A+P) + (B+Q) + (180 + C+R) = 180...
so we can say....(A+P) , (B+Q) , (180 + C+R) are three angles of a triangle....
similarly....rearranging them again we will get more SUCH equations....from which we can ssay...... (A+P) , (B+R) , (180 + C+Q) are three angles of a triangle....
and also....... (A+Q) , (B+P) , (180 + C+R) are three angles of a triangle....
and also....... (A+R) , (B+P) , (180 + C+Q) are three angles of a triangle....
and also....... (A+Q) , (B+R) , (180 + C+P) are three angles of a triangle....
and also....... (A+R) , (B+Q) , (180 + C+P) are three angles of a triangle....
PHEW.....the reason behind doing all this is comin....next.........
lets go back to the determinant....D.....
If we open it from the first row....then.....we will have.........
D = tan(A+P) { tan(B+Q)tan (C+R) - tan(C+Q)tan (B+R) } - tan (B+P) { tan(A+Q)tan(C+R) - tan(C+Q)tan(A+R) } + tan(C+P) { tan(A+Q)tan(B+R) - tan(B+Q)tan(A+R)}
now the main thing starts.......
when u open these brackets....u will get many terms of the form...
tan X tan Y tan Z........X,Y,Z may take values like-- (A+P), (B+Q) , (C+R) etc....
Now as we know.......in any triangle...say triangle XYZ....
X + Y + Z =180.....and for these angles.......
tanX tanY tanZ = tanX + tanY + tanZ
and using this property we can split those terms of multiplication form into addition.......and finally.....the opposite signs will cancel out.....
AND U WILL GET A BIG ZERO.........
hope its clear buddy.......