A) \[1\]
B) \[0\]
C) \[\cos A\cos B\operatorname{cosC}\]
D) \[\cos A+\cos B\,\cos \,C\]
Correct Answer: B
Solution :
| \[A+B+C=\pi ,\] therefore \[A+B=\pi -C\] or \[\cos \,(A+B)=\cos \,\left( \pi -C \right)=-\cos C\] |
| or \[\cos A\cos B-\sin A\sin B=-\cos C\]\[\Rightarrow \,\,\cos A\cos B+\cos C=\sin A\sin B\] and \[\sin \,(A+B)=\sin (\pi -C)=\sin C.\] |
| Expanding the given determinant, we get \[\Delta =-(1-{{\cos }^{2}}A)+\cos C(\cos C+\cos A+\cos B)\]\[+\cos \,B(\cos B+\cos A\cos C)\] |
| \[=-{{\sin }^{2}}A+\cos \,C(\sin A\sin B)+\cos B(\sin A\sin C)\] |
| \[=-{{\sin }^{2}}A+\sin A\,(\sin B\,\cos C+\cos B\sin C)\] |
| \[=-{{\sin }^{2}}A+\sin A\,\sin (B+C)=-{{\sin }^{2}}A+{{\sin }^{2}}A=0\] |
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