If in a triangle $ABC$, $b = ( a + c ) \cos \theta,$ find $\sin \theta$

Question

If in a triangle $ABC$, $b = ( a + c ) \cos \theta$, find $\sin \theta$ Please help, I wasn't able to figure out how to solve this.

Options : A) $\frac{2\sqrt{bc}}{b+c} \cos \frac{A}{2} $ B) $\frac{2\sqrt{ac}}{a+c} \cos \frac{B}{2} $ C) $\frac{2\sqrt{ab}}{a-b} \sin \frac{C}{2} $ D) $\frac{2\sqrt{bc}}{b-c} \sin \frac{A}{2} $

Answer 1

Using $$\cos(\theta)=\frac{b}{a+c}$$ we get by squaring $$\frac{b^2}{(b+c)^2}=\cos^2(\theta)=1-\sin^2(\theta)$$

Answer 2

$ \cos (\theta) =\frac {b}{a+c} $ $$ \cos (\theta) =\frac{(Sin B)}{(Sin A\,+ SinC )} $$ $$ \cos(\theta) = \frac{ Sin (B/2) }{Cos (( A -C)/2)} $$ $$ \cos (\theta) =\frac { Cos( (A+C)/2) }{Cos (( A-C)/2)} $$ $$ \cos (\theta) = \frac {1-(tanA/2)(tanC/2)}{1+(tanA/2)(tanB/2)}$$