Question

In a trangle,
prove that c(acosB-bcosA)=a^2-b^2
here a,b,c are sides and A,B are coressponding angles.​

Answer 1

Answer:

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Answer 2

Answer:

$c(acosB-bcosA)=a^2-b^2$ is proved.

Step-by-step explanation:

We know that

 $cosB=\frac{a^+c^2-b^2}{2ac}$               and       $cosA=\frac{b^2+c^2-a^2}{2bc}$

Simplifying $c(acosB-bcosA)=a^2-b^2$

L.H.S $\implies$  $c(a \hspace{0.1cm}\frac{a^+c^2-b^2}{2ac} - b \hspace{0.1cm} \frac{b^2+c^2-a^2}{2bc})$

         $\implies$   $c( \hspace{0.1cm}\frac{a^+c^2-b^2}{2c} - \hspace{0.1cm} \frac{b^2+c^2-a^2}{2c})$

         $\implies$    $c( \hspace{0.1cm}\frac{(a^+c^2-b^2)\hspace{0.1cm} -\hspace{0.1cm} (b^2+c^2-a^2)}{2c} )$

         $\implies$     $( \hspace{0.1cm}\frac{a^+c^2-b^2\hspace{0.1cm} -\hspace{0.1cm} b^2-c^2+a^2}{2} )$

         $\implies$     $( \hspace{0.1cm}\frac{2a^{2} \hspace{0.1cm}-\hspace{0.1cm}2b^2}{2} )$

         $\implies$     $a^2-b^2$

Hence Proved.

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