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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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Verified Answer
The correct answer is:
$2 \sqrt{c a} \sin \frac{B}{2}$
We have, $b \cos \theta=c-a \Rightarrow \cos \theta=\frac{c-a}{b}$
$$
\begin{aligned}
& \therefore \quad \sin \theta=\frac{\sqrt{b^2-(c-a)^2}}{b} \\
& \text { and } \tan \theta=\frac{\sqrt{b^2-(c-a)^2}}{(c-a)}
\end{aligned}
$$
and $\tan \theta=\frac{\sqrt{b^2-(c-a)^2}}{(c-a)}$
Now, $\quad(c-a) \tan \theta=\sqrt{b^2-(c-a)^2}$
$$
\begin{aligned}
& =\sqrt{b^2-c^2-a^2+2 a c}=\sqrt{-\left(c^2+a^2-b^2\right)+2 a c} \\
& =\sqrt{2 a c-2 a c \cos B} \quad \quad \text {[ Using cosine rule] } \\
& =\sqrt{2 a c} \sqrt{1-\cos B} \\
& =\sqrt{2} \sqrt{a c} \sqrt{2 \sin ^2 \frac{B}{2}}=2 \sqrt{a c} \sin \frac{B}{2}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \quad \sin \theta=\frac{\sqrt{b^2-(c-a)^2}}{b} \\
& \text { and } \tan \theta=\frac{\sqrt{b^2-(c-a)^2}}{(c-a)}
\end{aligned}
$$
and $\tan \theta=\frac{\sqrt{b^2-(c-a)^2}}{(c-a)}$
Now, $\quad(c-a) \tan \theta=\sqrt{b^2-(c-a)^2}$
$$
\begin{aligned}
& =\sqrt{b^2-c^2-a^2+2 a c}=\sqrt{-\left(c^2+a^2-b^2\right)+2 a c} \\
& =\sqrt{2 a c-2 a c \cos B} \quad \quad \text {[ Using cosine rule] } \\
& =\sqrt{2 a c} \sqrt{1-\cos B} \\
& =\sqrt{2} \sqrt{a c} \sqrt{2 \sin ^2 \frac{B}{2}}=2 \sqrt{a c} \sin \frac{B}{2}
\end{aligned}
$$
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