Prove- c-bcos A b-ccos A - cos Bcos C

We have to prove that, c bcos Ab ccos A=cos Bcos C We know, Cosine rule #160; cos A=b^2+c^2 a^22bc cos B=a^2+c^2 b^2ac cos C=b^2+a^2 c^22ab ...

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Standard XII

Mathematics

Cosine Rule

Prove: c-bco...

Question

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

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Solution

We have to prove that, $\frac{c - b cos A}{b - c cos A} = \frac{cos B}{cos C}$

We know, Cosine rule
$cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$

$cos B = \frac{a^{2} + c^{2} - b^{2}}{a c}$

$cos C = \frac{b^{2} + a^{2} - c^{2}}{2 a b}$

LHS
$c - b cos A = \frac{2 c^{2} - b^{2} - c^{2} + a^{2}}{2 c} = \frac{a^{2} + c^{2} - b^{2}}{2 c}$

$b - c cos A = \frac{2 b^{2} - c^{2} - b^{2} + a^{2}}{2 b} = \frac{a^{2} + b^{2} - c^{2}}{2 b}$

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

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

LHS $=$ RHS.

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Prove:c−bcos Ab−ccos A=cos Bcos C

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