A) 1
B) 0
C) \[\cos A\cos B\cos C\]
D) \[\cos A+\cos B\cos C\]
Correct Answer: B
Solution :
Given, Angles of a triangle = A, B and C. We know that as A + B + C = p, therefore \[A+B=\pi -C\] or \[\cos (A+B)=\cos (\pi -C)=-\cos C\] or \[\cos A\cos B-\sin A\sin B=-\cos C\] \[\cos A\cos B+\cos C=\sin A\sin B\] and \[\sin (A+B)=\sin (\pi -C)=\sin C.\] Expanding the given determinant, we get \[\Delta =-(1-{{\cos }^{2}}A)+\cos C(\cos C+\cos A\cos B)\] \[+\cos B(\cos B+\cos A\cos C)\] \[=-{{\sin }^{2}}A+\cos C(\sin A\sin B)+\cos B(\sin A\sin C)\] \[=-{{\sin }^{2}}A+\sin A(\sin B\cos C+\cos B\sin C)\] \[=-{{\sin }^{2}}A+\sin A\sin (B+C)\]\[=-{{\sin }^{2}}A+{{\sin }^{2}}A=0.\]You need to login to perform this action.
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