Question

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

• A. Area of $△ABC$
• B. Perimeter of $△ABC$
• C. $\Delta =\sum a^2$
• D. None of these

Answer 1

The correct option is D None of these

$\Delta =\left|\begin{array}{ccc}{a}^2& b\sin A& c\sin A\\ b\sin A& 1& \cos A\\ c\sin A& \cos A& 1\end{array}\right|$

Using Sine law $\sin A=ak,\sin B=bk$ and $\sin C=ck$

$\Delta =\left|\begin{array}{ccc}{a}^2& bak& cak\\ bak& 1& \cos A\\ cak& \cos A& 1\end{array}\right|=a^2\left|\begin{array}{ccc}1& bk& ck\\ bk& 1& \cos A\\ ck& \cos A& 1\end{array}\right|=a^2\left|\begin{array}{ccc}1& \sin B& \sin C\\ \sin B& 1& \cos A\\ \sin C& \cos A& 1\end{array}\right|$

Applying $C_2\to C_2-\sin B{C}_1,C_3\to C_3-\sin C{C}_1$

$\Delta =a^2\left|\begin{array}{ccc}1& 0& 0\\ \sin B& 1-{\sin }^{2}B& \cos A-\sin B\sin C\\ \sin C& \cos A-\sin B\sin C& 1-{\sin }^{2}C\end{array}\right|$

$\Delta =a^2({\cos }^{2}B{\cos }^{2}C-{(\cos A-\sin B\sin C)}^2)$

Now

$\cos A=\cos (\pi -(B+C))=-\cos (B+C)=-(\cos B\cos C-\sin B\sin C)$

$\Rightarrow \cos A-\sin B\sin C=-\cos B\cos C$

Substituting this we get

$\Delta =0$