In any ΔABC, prove that

Question:

In any ΔABC, prove that

$\frac{c-b \cos A}{b-\cos A}=\frac{\cos B}{\cos C}$

 

Solution:

Need to prove: $\frac{\mathrm{c}-\mathrm{b} \cos \mathrm{A}}{\mathrm{b}-\mathrm{c} \cos \mathrm{A}}=\frac{\cos \mathrm{B}}{\cos \mathrm{C}}$

Left hand side

$=\frac{c-b \cos A}{b-c \cos A}$

$=\frac{c-b \frac{b^{2}+c^{2}-a^{2}}{2 b c}}{b-c \frac{b^{2}+c^{2}-a^{2}}{2 b c}}$

$=\frac{\frac{2 c^{2}-b^{2}-c^{2}+a^{2}}{2 c}}{\frac{2 b^{2}-b^{2}-c^{2}+a^{2}}{2 b}}$

$=\frac{\frac{c^{2}+a^{2}-b^{2}}{2 c}}{\frac{b^{2}+a^{2}-c^{2}}{2 b}}$

$=\frac{\frac{c^{2}+a^{2}-b^{2}}{2 a c}}{\frac{b^{2}+a^{2}-c^{2}}{2 a b}}\left[\right.$ Multiplying the numerator and denominator by $\left.\frac{1}{a}\right]$

$=\frac{\cos \mathrm{B}}{\cos \mathrm{C}}$

$=$ Right hand side. [Proved]

 

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