Question

In triangle $ABC$, the value of ${\sin }^{2}A-\cos B(\cos A\cos C+\cos B)-\cos C(\cos A$ $\cos B+\cos C)$ is

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (A)

${\sin }^{2}A-\cos B(\cos A\cos C+\cos B)-\cos C(\cos A\cos B+\cos C)$
$={\sin }^{2}A-{\cos }^{2}B-{\cos }^{2}C-2\cos A\cos B\cos C$
$={\sin }^{2}A-{\cos }^{2}B-{\cos }^{2}C-\cos C(\cos (A+B)+\cos (A-B))$
$={\sin }^{2}A-{\cos }^{2}B-{\cos }^{2}C-\cos C(\cos (\pi -C)+\cos (A-B))$
$={\sin }^{2}A-{\cos }^{2}B-\cos C\cos (A-B)$
$={\sin }^{2}A-{\cos }^{2}B+\cos (A+B)\cos (A-B)$
$={\sin }^{2}A-{\cos }^{2}B+{\cos }^{2}A-{\sin }^{2}B=1-1=0$