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If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are the angles of a $\Delta \mathrm{ABC}$, then with usual notations, $\frac{c^{2}-a^{2}+b^{2}}{a^{2}-b^{2}+c^{2}}=$
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Verified Answer
The correct answer is:
$\frac{\tan B}{\tan A}$
$\begin{array}{l}
\frac{c^{2}-a^{2}+b^{2}}{a^{2}-b^{2}+c^{2}} \\
=\frac{b^{2}+c^{2}-a^{2}}{a^{2}+c^{2}-b^{2}}
\end{array}$
Dividing numerator and denominator by $2 \mathrm{abc}$.
$\begin{array}{l}
=\frac{\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right) \times \frac{1}{a}}{\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right) \times \frac{1}{b}}=\frac{(\cos A)\left(\frac{1}{a}\right)}{(\cos B) \times \frac{1}{b}} \\
\end{array}$
$=\frac{b}{a} \times \frac{\cos A}{\cos B}=\frac{K \sin B}{K \sin A} \times \frac{\cos A}{\cos B}$ ...(by Sine rule)
$=\frac{\tan B}{\tan A}$
\frac{c^{2}-a^{2}+b^{2}}{a^{2}-b^{2}+c^{2}} \\
=\frac{b^{2}+c^{2}-a^{2}}{a^{2}+c^{2}-b^{2}}
\end{array}$
Dividing numerator and denominator by $2 \mathrm{abc}$.
$\begin{array}{l}
=\frac{\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right) \times \frac{1}{a}}{\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right) \times \frac{1}{b}}=\frac{(\cos A)\left(\frac{1}{a}\right)}{(\cos B) \times \frac{1}{b}} \\
\end{array}$
$=\frac{b}{a} \times \frac{\cos A}{\cos B}=\frac{K \sin B}{K \sin A} \times \frac{\cos A}{\cos B}$ ...(by Sine rule)
$=\frac{\tan B}{\tan A}$
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