Question

In $\Delta ABC,\sum a^2({\cos }^{2}B-{\cos }^{2}C)=$

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

$\sum a^2({\cos }^{2}B-{\cos }^{2}C)$

$=a^2({\cos }^{2}B-{\cos }^{2}C)+b^2({\cos }^{2}C-{\cos }^{2}A)+c^2({\cos }^{2}A-{\cos }^{2}B)$

$=a^2(1-{\sin }^{2}B-1+{\sin }^{2}C)+b^2(1-{\sin }^{2}C-1+{\sin }^{2}A)+c^2(1-{\sin }^{2}A-1+{\sin }^{2}B)$

$=a^2(-{\sin }^{2}B+{\sin }^{2}C)+b^2(-{\sin }^{2}C+{\sin }^{2}A)+c^2(-{\sin }^{2}A+{\sin }^{2}B)$

$=a^2({\sin }^{2}C-{\sin }^{2}B)+b^2({\sin }^{2}A-{\sin }^{2}C)+c^2({\sin }^{2}B-{\sin }^{2}A)$

Using Sine rule $\sin A=\frac{a}{2R},\sin B=\frac{b}{2R},\sin C=\frac{c}{2R}$

$=a^2(\frac{c^2}{4R^2}-\frac{b^2}{4R^2})+b^2(\frac{a^2}{4R^2}-\frac{c^2}{4R^2})+c^2(\frac{b^2}{4R^2}-\frac{a^2}{4R^2})$

$=\frac{1}{4R^2}(a^2{c}^2-a^2{b}^2)+\frac{1}{4R^2}(b^2{a}^2-b^2{c}^2)+\frac{1}{4R^2}(c^2{b}^2-c^2{a}^2)$

$=\frac{1}{4R^2}(a^2{c}^2-a^2{b}^2+b^2{a}^2-b^2{c}^2+c^2{b}^2-c^2{a}^2)$

$=\frac{1}{4R^2}\left(0\right)=0$