In ABC, prove that a ( C - B )=2( b - c ) ^2 (A 2 ) - Vidyadip

Question

In $\triangle ABC$, prove that $a (\cos C -\cos B )=2( b - c ) \cos ^2\left(\frac{A}{2}\right)$

✓

Answer

By Projection rule, we have $a \cos C + c \cos A = b$ and $a \cos B + b \cos A = c$

$

\begin{aligned}

\therefore \quad & a \cos C=b-c \cos A \text { and } a \cos B=c-b \cos A \\

& \text { L.H.S. }=a(\cos C-\cos B) \\

& =a \cos C-a \cos B \\

& =(b-c \cos A)-(c-b \cos A) \\

& =b-c \cos A-c+b \cos A \\

& =(b-c)+(b-c) \cos A \\

& =(b-c)(1+\cos A) \\

& =(b-c) \times 2 \cos ^2 \frac{A}{2} \\

& =2(b-c) \cos ^2 \frac{A}{2} \\

& =\text { R.H.S. }

\end{aligned}

$

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