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If $\theta$ does not lie in the second quadrant and $\theta=\frac{-3}{4}$, then $\tan \frac{\theta}{2}+\sin 2 \theta=$
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The correct answer is:
$\frac{-97}{75}$
Given $\tan \theta=\frac{-3}{4}$
$\begin{aligned} & \Rightarrow \sin \theta=\frac{-3}{5}, \cos \theta=\frac{4}{5} \Rightarrow \cos 2 \theta=2 \times \frac{-3}{5} \times \frac{4}{5}=\frac{-24}{25} \\ & \tan \frac{\theta}{2}=\frac{-1}{3} \Rightarrow \tan \frac{\theta}{2}+\sin 2 \theta=\frac{-1}{3}-\frac{24}{25}=\frac{-97}{75}\end{aligned}$
$\begin{aligned} & \Rightarrow \sin \theta=\frac{-3}{5}, \cos \theta=\frac{4}{5} \Rightarrow \cos 2 \theta=2 \times \frac{-3}{5} \times \frac{4}{5}=\frac{-24}{25} \\ & \tan \frac{\theta}{2}=\frac{-1}{3} \Rightarrow \tan \frac{\theta}{2}+\sin 2 \theta=\frac{-1}{3}-\frac{24}{25}=\frac{-97}{75}\end{aligned}$
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