Let S = θ ∈ [ - π , π ] - ± π 2 : sin θ tan θ + tan θ = sin 2…

Given $\sin \theta \tan \theta +\tan \theta =\sin 2\theta ; \theta \ne \pm \frac{\pi }{2}$

$\Rightarrow \frac{{\sin }^2 \theta }{\cos \theta }+\frac{\sin \theta }{\cos \theta }=2 \sin \theta \cos \theta$

$\Rightarrow \sin \theta =0$ or $\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }=2 \cos \theta$

$\Rightarrow \theta =0, \mathrm{\pi }, -\mathrm{\pi }$ or $\sin \theta +1=2\left(1-{\sin }^2 \theta \right)$

$\Rightarrow \theta =0, \pi , -\pi$ or $2 {\sin }^2 \theta +\sin \theta -1=0$

$\Rightarrow \theta =0, \pi , -\pi$ or $\sin \theta =\frac{1}{2}, \sin \theta =-1$

$\Rightarrow \theta =0, \pi , -\pi$ or $\theta =\frac{\pi }{6}, \frac{5\pi }{6}$ or $\theta =\frac{-\pi }{2}$ (rejected)

Hence, $\theta =\left\{0, \pi , -\pi , \frac{\pi }{6}, \frac{5\pi }{6}\right\}$

$\Rightarrow n\left(S\right)=5$

Now, $T=\Sigma \cos 2\theta$

$=\cos 0+\cos 2\pi +\cos \left(-2\pi \right)+\cos \left(\frac{\pi }{3}\right)+\cos \left(\frac{5\mathrm{\pi }}{3}\right)$

$=1+1+1+\frac{1}{2}+\frac{1}{2}=4$

So, $T+n\left(S\right)=9$