Question

In a triangle $ABC, a[b \,\cos \,C - c\, \cos\, B]$ =

• A. $a^2$
• B. $b^2$
• C. $0$
• D. $b^{2}-c^{2}$

Answer 1

Answer: (D)

$a[b \cos C-c \cos B]$
$=(b \cos C+c \cos B)(b \cos C-c \cos B)$
$=b^{2} \cos ^{2} C-c^{2} \cos ^{2} B$
$=b^{2}\left(1-\sin ^{2} C\right)-c^{2}\left(1-\sin ^{2} B\right)$
$=b^{2}\left(1-\frac{c^{2}}{4 R^{2}}\right)-c^{2}\left(1-\frac{b^{2}}{4 R^{2}}\right)$
$\left(\because \sin C=\frac{C}{2 R}\right)$
$=b^{2}-\frac{b^{2} c^{2}}{4 R^{2}}-c^{2}+\frac{c^{2} b^{2}}{4 R^{2}}$
$=b^{2}-c^{2}$