In a $△ A B C$ , prove that : $b (a cos C - c cos A) = a^{2} - c^{2}$

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Solution

$L H S$

$= b (a cos C - c cos A) = b (a \frac{a^{2} + b^{2} - c^{2}}{2 a b} - c \frac{b^{2} + c^{2} - a^{2}}{2 b c})$
$= \frac{b}{2 b} (a^{2} + b^{2} - c^{2} - b^{2} - c^{2} + a^{2}) = \frac{1}{2} (2 a^{2} - 2 c^{2}) = a^{2} - c^{2}$
$= R H S$

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In a △ABC, prove that : b(acosC−ccosA)=a2−c2

LHS=b(acosC−ccosA)=b(aa2+b2−c22ab−cb2+c2−a22bc)=b2b(a2+b2−c2−b2−c2+a2)=12(2a2−2c2)=a2−c2=RHS

In a $△ A B C$ , prove that : $b (a cos C - c cos A) = a^{2} - c^{2}$

$L H S$

$= b (a cos C - c cos A) = b (a \frac{a^{2} + b^{2} - c^{2}}{2 a b} - c \frac{b^{2} + c^{2} - a^{2}}{2 b c})$
$= \frac{b}{2 b} (a^{2} + b^{2} - c^{2} - b^{2} - c^{2} + a^{2}) = \frac{1}{2} (2 a^{2} - 2 c^{2}) = a^{2} - c^{2}$
$= R H S$