Math, asked by prabinsingh9816, 1 month ago

(c) sin (A - B) =à²-b²/c² (sinc) ​

Answers

Answered by mathdude500
11

Appropriate Question ;- In  \triangleABC, 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 \: }} \\  \\

Answered by 9218
1

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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