In $△ A B C$ , $2 a (b cos C - c cos B) =$ $b^{2} - c^{2}$ . Is it true ? If true enter $1$ else $0$ .

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Solution

We know that the cosine rule
$b^{2} = a^{2} + c^{2} - 2 a c cos B$ $. . . . . . . . . . . . (1)$

$c^{2} = a^{2} + b^{2} - 2 a b cos C$ $. . . . . . . . . . (2)$

From $(1) - (2)$ , we get
$b^{2} - c^{2} = a^{2} + c^{2} - 2 a c cos B - a^{2} - c^{2} + 2 a b cos C$
$b^{2} - c^{2} = 2 a b cos C - 2 a c cos B$
$b^{2} - c^{2} = 2 a (b cos C - c cos B)$

So, it is true.

Therefore, the answer will be $1$ .

Hence, this is the answer.

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