Question

If a, b and c are the sides of a triangle and A, B and C are the angles opposite to a, b and c respectively, then
$\Delta =\left|\begin{array}{ccc}{a}^2& b\sin A& c\sin A\\ b\sin A& 1& \cos A\\ c\sin A& \cos A& 1\end{array}\right|$ is independent of

• A. a
• B. b
• C. c
• D. A, B, C

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A a
B b
C c
D A, B, C

$\Delta =\left|\begin{array}{ccc}{a}^2& b\sin A& c\sin A\\ b\sin A& 1& \cos A\\ c\sin A& \cos A& 1\end{array}\right|$
Using Sin 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
$\Delta =0$