Question

If $a, b, c$ are the sides of a $\triangle A B C$ opposite angles $A, B, C$ respectively, and $\Delta=\begin{vmatrix}a^2 & b \sin A & c \sin A \\ b \sin A & 1 & \cos (B-C) \\ c \sin A & \cos (B-C) & 1\end{vmatrix}$, then $\Delta$ equals

• A. $\sin A-\sin C \sin B$
• B. $a b c$
• C. 1
• D. 0

Answer 1

Answer: (D)

By the law of sines
$\frac{a}{\sin A} =\frac{b}{\sin B}=\frac{c}{\sin C}=k \text { (say) } $
$\Rightarrow a=k \sin A, b =k \sin B, c=k \sin C \text {. Now } $
$\Delta= \begin{vmatrix}a^2 & a b / k & a c / k \\a b / k & 1 & \cos (B-C) \\a c / k & \cos (B-C) & 1\end{vmatrix} $
$= a^2\begin{vmatrix}1 & \sin B & \sin C \\\sin B & 1 & \cos (B-C) \\\sin C & \cos (B-C) & 1\end{vmatrix}$
$= a^2\begin{vmatrix}1 & \sin (A+C) & \sin (A+B) \\\sin (A+C) & 1 & \cos (B-C) \\\sin (A+B) & \cos (B-C) & 1\end{vmatrix}$
$= a^2\begin{vmatrix}\sin A & \cos A & 0 \\\cos C & \sin C & 0 \\\cos B &\sin B & 0\end{vmatrix} \begin{vmatrix}
\sin A & \cos A & 0 \\\cos C & \sin C & 0 \\\cos B & \sin B & 0\end{vmatrix}
$=a^2(0)=0$