In triangle ABC,if b cos theta = c - a,(where | vedclass

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(B) Given $b \cos 	heta = c - a$,we have $\cos 	heta = \frac{c - a}{b}$.Since $	heta$ is an acute angle,$\sin 	heta = \sqrt{1 - \cos^2 	heta} = \

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$2 \sqrt{ca} \sin \frac{B}{2}$

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(B) Given $b \cos 	heta = c - a$,we have $\cos 	heta = \frac{c - a}{b}$.Since $	heta$ is an acute angle,$\sin 	heta = \sqrt{1 - \cos^2 	heta} = \frac{\sqrt{b^2 - (c - a)^2}}{b}$.Thus,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\sqrt{b^2 - (c - a)^2}}{c - a}$.Therefore,$(c - a) 	an 	heta = \sqrt{b^2 - (c - a)^2} = \sqrt{b^2 - c^2 - a^2 + 2ac}$.Using the cosine rule $b^2 = a^2 + c^2 - 2ac \cos B$,we substitute $b^2 - c^2 - a^2 = -2ac \cos B$.So,$(c - a) 	an 	heta = \sqrt{2ac - 2ac \cos B} = \sqrt{2ac(1 - \cos B)}$.Using the identity $1 - \cos B = 2 \sin^2 \frac{B}{2}$,we get:$(c - a) 	an 	heta = \sqrt{2ac \cdot 2 \sin^2 \frac{B}{2}} = \sqrt{4ac \sin^2 \frac{B}{2}} = 2 \sqrt{ca} \sin \frac{B}{2}$.

In $\triangle ABC$,if $b \cos \theta = c - a$,(where $\theta$ is an acute angle),then $(c - a) \tan \theta =$

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