Question

If $A + B + C = \pi ,$then $\left| {\begin{array}{*{20}{c}}
  {\sin (A + B + C)}&{\sin B}&{\cos C} \\
  {\sin B}&0&{\tan A} \\
  {\cos (A + B)}&{ - \tan A}&0
\end{array}} \right| = .......................$

Answer 1

Hint: $\sin \pi = 0$, $\cos (\pi - x) = - \cos x$
Using the properties of trigonometry first, we make the determinant simple and then further solve it.
Also given
$A + B + C = \pi$
Applying trigonometric properties, we get
$
  \sin (A + B + C) = \sin \pi = 0 \\
  \cos (A + B) = \cos (\pi - C) = - \cos C \\
$
Putting these values in the determinant, we get
$
  \Delta = \left| {\begin{array}{*{20}{c}}
  {\sin (A + B + C)}&{\sin B}&{\cos C} \\
  {\sin B}&0&{\tan A} \\
  {\cos (A + B)}&{ - \tan A}&0
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  0&{\sin B}&{\cos C} \\
  {\sin B}&0&{\tan A} \\
  { - \cos C}&{ - \tan A}&0
\end{array}} \right| \\
    \\
$
Now expanding the determinant through row${R_1}$,we get
$\begin{gathered}
  \Delta = 0\left| {\begin{array}{*{20}{c}}
  0&{\tan A} \\
  { - \tan A}&0
\end{array}} \right| + ( - \sin B)\left| {\begin{array}{*{20}{c}}
  {\sin B}&{\tan A} \\
  { - \cos C}&0
\end{array}} \right| + \cos C\left| {\begin{array}{*{20}{c}}
  {\sin B}&0 \\
  { - \cos C}&{ - \tan A}
\end{array}} \right| \\
  \Delta = - \sin B(0 - (\tan A)( - \cos C)) + \cos C((\sin B)( - \tan A) - 0) \\
  \Delta = \sin B\tan A\cos C - \sin B\tan A\cos C \\
  \Delta = 0 \\
\end{gathered} $
Therefore, value of determinant,
$\Delta = \left| {\begin{array}{*{20}{c}}
  {\sin (A + B + C)}&{\sin B}&{\cos C} \\
  {\sin B}&0&{\tan A} \\
  {\cos (A + B)}&{ - \tan A}&0
\end{array}} \right| = 0$
Note: Use of trigonometric properties transforms complex determinant into simple determinant
So, first apply trigonometric properties and then expand the determinant. As this will be the
easiest and efficient way to get the solution of such problems.