Question

If $a,b,c$ are the sides of a $△ABC$ opposite angles $A,B,C$ respectively, and $\Delta =\left|\begin{array}{ccc}{a}^2& b\sin A& c\sin A\\ b\sin A& 1& \cos (B-C)\\ c\sin A& \cos (B-C)& 1\end{array}\right|$, then $\Delta$ equals

• A. $\sin A-\sin C\sin B$
• B. $abc$
• C. 1
• D. 0

Answer 1

Answer: (D)

By the law of sines
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\text{ (say) }$
$\Rightarrow a=k\sin A,b=k\sin B,c=k\sin C\text{. Now }$
$\Delta =\left|\begin{array}{ccc}{a}^2& ab/k& ac/k\\ ab/k& 1& \cos (B-C)\\ ac/k& \cos (B-C)& 1\end{array}\right|$
$=a^2\left|\begin{array}{ccc}1& \sin B& \sin C\\ \sin B& 1& \cos (B-C)\\ \sin C& \cos (B-C)& 1\end{array}\right|$
$=a^2\left|\begin{array}{ccc}1& \sin (A+C)& \sin (A+B)\\ \sin (A+C)& 1& \cos (B-C)\\ \sin (A+B)& \cos (B-C)& 1\end{array}\right|$
$=a^2\left|\begin{array}{ccc}\sin A& \cos A& 0\\ \cos C& \sin C& 0\\ \cos B& \sin B& 0\end{array}\right|\left|\begin{array}{ccc}<br/>\sin A& \cos A& 0\\ \cos C& \sin C& 0\\ \cos B& \sin B& 0\end{array}\right|<br/>$=a^2(0)=0$