Question

In ΔABC, show that $\frac{b^{2}\ -\ c^{2}}{a^{2}}\ =\ \frac{sin(B - C)}{sin(B + C)}$.

Answer 1

Answer:

Step-by-step explanation:

Formula used:

1. Sine formula:

a/sinA = b/sinB = c/sinC = 2R

2. sin²A - sin²B = sin(A+B) sin(A-B)

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

$=\frac{{(2R\:sinB)}^2-{(2R\:sinB)}^2}{{2R\:sinC}^2}$

$=\frac{{(2R)}^{2}({ sin }^{2}B-{sin}^{2}C)}{{(2r)}^{2}{sin}^{2}A}$

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

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

$=\frac{sin(B+C)\:sin(B-C)}{ {sin}^{2}(B + C)}$

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

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