Question

In any triangle ABC ((b^2-c^2)/a^2) is equal to​

Answer 1

Basic Identities Used :-

$1. \:\red{\bf \:Sine \: law}$

In triangle ABC, if a, b and c represents the sides BC, CA and AB respectively, then

$\sf \: \dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC} = k$

$\sf :\longmapsto\:a = k \: sinA$

$\sf :\longmapsto\:b = k \: sinB$

$\sf :\longmapsto\:c = k \: sinC$

$2. \: \red{ \bf \: {sin}^{2}x - {sin}^{2}y = sin(x + y)sin(x - y)}$

$3. \: \red{ \bf \: sin(\pi \: - x) = sinx}$

Let's solve the problem now!!

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\:\dfrac{ {b}^{2} - {c}^{2} }{ {a}^{2} }$

On substituting the values of a, b and c from sine law, we get

$\rm \:  \:  =  \:  \: \dfrac{ {(ksinB)}^{2} - {(ksinC)}^{2} }{ {(ksinA)}^{2} }$

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

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

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

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

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

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \blue{ \bf \: A + B + C = \pi \}}}$

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

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

Additional Information :-

Cosine Law :-

In triangle ABC, if a, b and c represents the sides BC, CA and AB respectively, then

$\boxed{ \purple{ \sf \: cosA = \dfrac{ {b}^{2} + {c}^{2} - {a}^{2}}{2bc}}}$

$\boxed{ \purple{ \sf \: cosB = \dfrac{ {c}^{2} + {a}^{2} - {b}^{2}}{2ac}}}$

$\boxed{ \purple{ \sf \: cosC = \dfrac{ {a}^{2} + {b}^{2} - {c}^{2}}{2ab}}}$

Projection Formulas :-

$\boxed{ \purple{\sf \: \: \: a \: = \: b \: cosC \: + \: c \: cosB}}$

$\boxed{ \purple{\sf \: \: \: b \: = \: c \: cosA \: + \: a \: cosC}}$

$\boxed{ \purple{\sf \: \: \: c \: = \: a \: cos B\: + \: b \: cosA}}$