Question

In a triangle $ABC$ with usual notation, which of the following is (are) CORRECT?

• A. If the angles $A,B,C$ are in A.P., then $b^2=c^2+a^2-ca$
• B. $\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^2-c^2}{a^2}$
• C. $\sum \frac{b^2-c^2}{\cos B+\cos C}=0$
• D. $\frac{1+\cos (A-B)\cos C}{1+\cos (A-C)\cos B}=\frac{a^2+b^2}{a^2+c^2}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A If the angles $A,B,C$ are in A.P., then $b^2=c^2+a^2-ca$
B $\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^2-c^2}{a^2}$
C $\sum \frac{b^2-c^2}{\cos B+\cos C}=0$
D $\frac{1+\cos (A-B)\cos C}{1+\cos (A-C)\cos B}=\frac{a^2+b^2}{a^2+c^2}$

$A,B,C$ are in A.P.
$\Rightarrow B={60}^{\circ }$
Now, by cosine rule,
$b^2=c^2+a^2-2ca\cos B$
$=c^2+a^2-2ca\times \frac{1}{2}$
$=c^2+a^2-2ca$

$\frac{\sin (B-C)}{\sin (B+C)}$
$=\frac{\sin (B-C)\sin (B+C)}{{\sin }^2(B+C)}$
$=\frac{{\sin }^{2}B-{\sin }^{2}C}{{\sin }^{2}A}$
$=\frac{b^2-c^2}{a^2}$ [By sine rule]

$\frac{b^2-c^2}{\cos B+\cos C}$
$=\frac{4R^2({\sin }^{2}B-{\sin }^{2}C)}{\cos B+\cos C}$ [By sine rule]
$=\frac{4R^2({\cos }^{2}C-{\cos }^{2}B)}{\cos B+\cos C}$
$=4R^2(\cos C-\cos B)$
$\therefore \sum \frac{b^2-c^2}{\cos B+\cos C}$
$=4R^2(\cos C-\cos B+\cos A-\cos C+\cos B-\cos A)$
$=0$

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