In ∆ABC, prove the following:

Question:

In ∆ABC, prove the following:

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

Solution:

Let ABC be any triangle.

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

$=\frac{a \cos B+b \cos A-b \cos A}{a \cos C+c \cos A-c \cos A} \quad$ [Using projection formula :

$c=a \cos B+b \cos A, b=a \cos C+c \cos A]$

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

 

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

Hence proved.

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