Question

Prove that in a triangle with angles A,B,C and opposite sides as a,b,c $\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^2-c^2}{a^2}$

Answer 1

To prove $\to \frac{\sin (B-C)}{\sin (B+C)}\frac{b^2-c^2}{a^2}$

where $A,B,C,\to$ angles of $△$

$a,b,c\to$ sides of $△$

defined such that,

where $A+B+C=\pi$

Using Sine Rule for a $△$

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=b\left(let\right)$

From here, we get

$a=\frac{\sin A}{b},b=\frac{\sin B}{b},c=\frac{\sin C}{b}$

$\Rightarrow$ Put values $a,b,c$ on $RHS$

$RHS=\frac{b^2-c^2}{a^2}=\frac{\frac{{\sin }^{2}B}{b^2}-\frac{{\sin }^{2}C}{b^2}}{\frac{{\sin }^{2}A}{b^2}}$

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

$[UsingFormula\to \sin (B+C\left)\sin \right(B-C)={\sin }^{2}B-{\sin }^{2}C]$

$=\frac{\sin (B+C)\sin (B-C)}{{\sin }^{2}A}[AsA+B+C=\pi A=\pi -(B+C\left)\right]$

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

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

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

Hence proved.