let $\tan \frac{\theta}{2}=\mathrm{t}$
$ \frac{4-4 t^2-6 t}{1+t^2}=1 $
$ 4-4 t^2-6 t=1+t^2$
$ \Rightarrow 5 t^2+6 t-3=0 $
$ \Rightarrow t=\frac{-6 \pm \sqrt{36-4(5)(-3)}}{2(5)} $
$ =\frac{-6 \pm \sqrt{96}}{10} $
$ =\frac{-6 \pm 4 \sqrt{6}}{10} $
$ t=\frac{-3+2 \sqrt{6}}{5} $
$ \cos \theta=\frac{1-t^2}{1+t^2}=\frac{1-\left(\frac{2 \sqrt{6}-3}{5}\right)^2}{1+\left(\frac{2 \sqrt{6}-3}{5}\right)^2}=\frac{1-\left(\frac{24+9-12 \sqrt{6}}{25}\right)}{1+\left(\frac{24+9-12 \sqrt{6}}{25}\right)} $
$ =\frac{25-33+12 \sqrt{6}}{25+33-12 \sqrt{6}}=\frac{12 \sqrt{6}-8}{58-12 \sqrt{6}}=\frac{6 \sqrt{6}-4}{29-6 \sqrt{6}} \times \frac{29+6 \sqrt{6}}{29+6 \sqrt{6}} $
$ =\frac{100+150 \sqrt{6}}{625}=\frac{4+6 \sqrt{6}}{25} \times \frac{4-6 \sqrt{6}}{4-6 \sqrt{6}} $
$ =\frac{-200}{25(4-6 \sqrt{6})}=\frac{-8}{4-6 \sqrt{6}}=\frac{4}{3 \sqrt{6}-2}$
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$f( x )=\frac{ x ^2-3 x -6}{ x ^2+2 x +4} \text {. }$
Then which of the following statements is (are) $TRUE$ ?
$(A)$ $f$ is decreasing in the interval $(-2,-1)$
$(B)$ $f$ is increasing in the interval $(1,2)$
$(C)$ $f$ is onto
$(D)$ Range of $f$ is $\left[-\frac{3}{2}, 2\right]$