Question

In any triangle ABC Prove that %_ sin(B-C)/sin(B+C)=(b^2 - c^2)/a^2

Answer 1

$\Large{\textbf{\underline{\underline{According\;to\;the\;Questions}}}}$

Here,

Assume

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}=Z$

a = ZsinA

b = ZsinB

c = ZsinC

Therefore,

RHS

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{(b^2-c^2)}{a^2}=\dfrac{Z^2 sin^2 B-Z^2 sin^2 C}{Z^2 sin^2 A}$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{Z^2 sin^2 B-Z^2 sin^2 C}{Z^2 sin^2 A}$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{(sin^2 B-sin^2 C)}{sin^2 A}$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{sin(B+C)\times sin(B-C)}{sin^2 (B+C)}$

We know that :-

(A + B + C) = π

A = π - (B + C)

Also,

sinA = sin[π - (B + C)] = sin(B + C)

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

Therefore,

LHS = RHS

Hence,

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