Question

(c) sin (A - B) =à²-b²/c² (sinc) ​

Answer 1

Appropriate Question ;- In $\triangle$ABC, prove that

$\sf \: sin(A - B) = \dfrac{ {a}^{2} - {b}^{2} }{ {c}^{2} }sinC$

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

Consider, RHS

$\sf \: sin(A - B) = \dfrac{ {a}^{2} - {b}^{2} }{ {c}^{2} }sinC$

We know, Sine Law

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

$\bf\implies \:a = ksinA \:$

$\bf\implies \:b = ksinB \:$

$\bf\implies \:c = ksinC \:$

So, on substituting these values in above expression, we get

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

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

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

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

We know,

$\boxed{ \sf{ \:\sf \: A + B + C = \pi \: \: }}$

So, using this, we get

$\sf \: = \: \dfrac{sin(\pi - C) \: sin(A - B)}{{sin} C} \:$

$\sf \: = \: \dfrac{sinC \: sin(A - B)}{{sin} C} \:$

$\sf \: = \: sin(A - B)$

Hence,

$\bf\implies \: sin(A - B) = \dfrac{ {a}^{2} - {b}^{2} }{ {c}^{2} }sinC$

$\rule{190pt}{2pt}$

Additional Information :-

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

$\boxed{ \sf{ \:cosB = \frac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ca} \: }}$

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

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

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

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

Answer 2

❒ Information provide to us ❒

In \triangle△ ABC, prove that

$\begin{gathered}\sf \: sin(A - B) = \dfrac{ {a}^{2} - {b}^{2} }{ {c}^{2} }sinC \\ \\ \end{gathered}$

❒ Solution ❒

Consider, RHS

$\begin{gathered}\sf \: sin(A - B) = \dfrac{ {a}^{2} - {b}^{2} }{ {c}^{2} }sinC \\ \\ \end{gathered}$

$\sf{{We \: know, \: Sine \: Law}}\\</p><p>\begin{gathered}\sf \: \dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC} = k \\ \end{gathered} \\ \begin{gathered}\bf\implies \:a = ksinA \: \end{gathered} \\</p><p>\begin{gathered}\bf\implies \:b = ksinB \: \end{gathered} \begin{gathered}\bf\implies \:c = ksinC \: \end{gathered}\\ \text{So, \: on \: substituting \:} \\ \text{these \: values \: in \: above } \: \\ \text{expression, \: we \: get} \\ \\ \begin{gathered}\sf \: = \: \dfrac{ {k}^{2} {sin}^{2} A - {k}^{2} {sin}^{2} B}{ {k}^{2} {sin}^{2} C} \: sinC \\ \end{gathered}$

$\begin{gathered}\sf \: = \: \dfrac{ {k}^{2} ({sin}^{2} A - {sin}^{2} B)}{ {k}^{2} {sin} C} \: \\ \\ \end{gathered} \\ \begin{gathered}\sf \: = \: \dfrac{{sin}^{2} A - {sin}^{2} B}{{sin} C} \: \\ \\ \end{gathered} \\ \begin{gathered}\sf \: = \: \dfrac{sin(A + B) \: sin(A - B)}{{sin} C} \: \\ \\ \end{gathered} \\ \sf{We \: know,} \\\begin{gathered}\boxed{ \sf{ \:\sf \: A + B + C = \pi \: \: }} \\ \\ \end{gathered} \\ \rm{So,\: using \:this, \:we \:get} \\ \begin{gathered}\sf \: = \: \dfrac{sin(\pi - C) \: sin(A - B)}{{sin} C} \: \\ \\ \end{gathered}$

$\begin{gathered}\sf \: = \: \dfrac{sinC \: sin(A - B)}{{sin} C} \: \\ \\ \end{gathered} </p><p>\begin{gathered}\sf \: = \: sin(A - B) \end{gathered} \underline{\sf{Hence}},</p><p>\begin{gathered}\bf\implies \: sin(A - B) = \dfrac{ {a}^{2} - {b}^{2} }{ {c}^{2} }sinC \end{gathered}$

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