In ∆ABC, prove the following:
$b (c \cos A - a \cos C) = c^{2} - a^{2}$

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Solution

Let ABC be any triangle.

$Consider b (c \cos A - a \cos C) = b c \cos A - a b \cos C = b c (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) - a b (\frac{a^{2} + b^{2} - c^{2}}{2 a b}) = \frac{b^{2} + c^{2} - a^{2} - a^{2} - b^{2} + c^{2}}{2} = \frac{2 (c^{2} - a^{2})}{2} = c^{2} - a^{2}$

Hence proved.

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Similar questions

$b (c c o s A - a c o s C) = c^{2} - a^{2}$

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\frac{cos 2 A}{a^{2}} - \frac{cos 2 C}{c^{2}} = \frac{1}{a^{2}} - \frac{1}{c^{2}}

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\frac{tan A}{tan B} = \frac{c^{2} + a^{2} - b^{2}}{c^{2} + b^{2} - a^{2}}

Q. In any triangle

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\frac{b^{2} - c^{2}}{a^{2}} sin 2 A + \frac{c^{2} - a^{2}}{b^{2}} sin 2 B + \frac{a^{2} - b^{2}}{c^{2}} sin 2 C = 0

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In ∆ABC, prove the following: b c cos A-a cos C=c2-a2

Let ABC be any triangle. Consider bccosA-acosC=bccosA-abcosC =bcb2+c2-a22bc-aba2+b2-c22ab =b2+c2-a2-a2-b2+c22 =2c2-a22 =c2-a2 Hence proved.

In ∆ABC, prove the following:
$b (c \cos A - a \cos C) = c^{2} - a^{2}$

Let ABC be any triangle.

$Consider b (c \cos A - a \cos C) = b c \cos A - a b \cos C = b c (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) - a b (\frac{a^{2} + b^{2} - c^{2}}{2 a b}) = \frac{b^{2} + c^{2} - a^{2} - a^{2} - b^{2} + c^{2}}{2} = \frac{2 (c^{2} - a^{2})}{2} = c^{2} - a^{2}$

Hence proved.