In $ riangle A B C$, if $b \cos heta=c-a$, (where $ heta$ is an acute angle), then $(c-a) an heta=$

Question

In $	riangle A B C$, if $b \cos 	heta=c-a$, (where $	heta$ is an acute angle), then $(c-a) 	an 	heta=$

Aicharya · Accepted Answer

We have, $b \cos 	heta=c-a \Rightarrow \cos 	heta=\frac{c-a}{b}$ $ \begin{aligned} & 	herefore \quad \sin 	heta=\frac{\sqrt{b^2-(c-a)^2}}{b} \ & 	ext { and } 	an 	heta=\frac{\sqrt{b^2-(c-a)^2}}{(c-a)} \end{aligned} $ and $	an 	heta=\frac{\sqrt{b^2-(c-a)^2}}{(c-a)}$ Now, $\quad(c-a) 	an 	heta=\sqrt{b^2-(c-a)^2}$ $ \begin{aligned} & =\sqrt{b^2-c^2-a^2+2 a c}=\sqrt{-\left(c^2+a^2-b^2ight)+2 a c} \ & =\sqrt{2 a c-2 a c \cos B} \quad \quad 	ext {[ Using cosine rule] } \ & =\sqrt{2 a c} \sqrt{1-\cos B} \ & =\sqrt{2} \sqrt{a c} \sqrt{2 \sin ^2 \frac{B}{2}}=2 \sqrt{a c} \sin \frac{B}{2} \end{aligned} $

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

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