Question

$In\Delta ABC,\frac{b^2-c^2}{a^2}=$

• A. $\frac{sin\left(B-C\right)}{\sin \left(B+C\right)}$
• B. $\frac{sin\left(B+C\right)}{\sin \left(B-C\right)}$
• C. $\frac{sin\left(B+C\right)}{\cos \left(B-C\right)}$
• D. $\frac{cos\left(B+C\right)}{\cos \left(B-C\right)}$

Answer 1

The correct option is A $\frac{sin\left(B-C\right)}{\sin \left(B+C\right)}$

$\frac{b^2-c^2}{a^2}$

By sine rule we have $a=2R\sin A,b=2R\sin B$ and $c=2R\sin C$

$=\frac{4R^2{\sin }^{2}B-4R^2{\sin }^{2}C}{4R^2{\sin }^{2}A}$

$=\frac{{\sin }^{2}B-{\sin }^{2}C}{{\sin }^{2}A}$

$=\frac{\sin (B+C)\sin (B-C)}{{\sin }^2(\pi -(B+C))}$ since ${\sin }^{2}A-{\sin }^{2}B=\sin (A+B)\sin (A-B)$

$=\frac{\sin (B+C)\sin (B-C)}{{\sin }^2(B+C)}$ since $A=180-(B+C)$

$=\frac{\sin (B-C)}{\sin (B+C)}$