Then general values of $\theta$ in inverse Trigo sum 
If $\displaystyle \theta = \tan^{-1}\bigg(2\tan^2 \theta\bigg)-\frac{1}{2}\sin^{-1}\bigg(\frac{3\sin 2 \theta}{5+4\cos 2 \theta}\bigg).$ Then general values of $\theta$ is 

Try: Let $\alpha =\tan^{-1}\bigg(2\tan^2 \theta\bigg)\Rightarrow \tan \alpha =2 \tan^2 \theta$
and $\displaystyle \beta = \frac{1}{2}\sin^{-1}\bigg(\frac{3\sin 2 \theta}{5+4\cos 2 \theta}\bigg)\Rightarrow \sin (2\beta) = \frac{3\sin 2 \theta}{5+4\cos 2 \theta} = 3\frac{2\tan \theta}{1+\tan^2 \theta}\cdot \frac{1}{5+4\bigg(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\bigg)}$
So $\displaystyle \sin (2\beta) = \frac{6\tan \theta}{9+\tan^2 \theta}.$
could some help me how to find range of $\theta$. Thanks

Hint: 
multiply the original equation by $2$, take the sine of both sides and remember that $$\cot = \frac 1\tan\\1+\tan^2 = \sec^2\\1 + \cot^2 = \csc^2$$

Added:
$$\theta = \tan^{-1}u - \frac 12 \sin^{-1}v\\2\theta = 2\tan^{-1}u -\sin^{-1}v\\\sin 2\theta = \sin\left(2\tan^{-1}u -\sin^{-1}v\right)\\\sin 2\theta = \sin(2\tan^{-1}u)\cos(\sin^{-1}v) - \cos(2\tan^{-1}u)\sin(\sin^{-1}v) $$
Now you just need to apply double angle formulas, then reduce the various "function of inverse function" combinations, which all have strictly algebraic expressions (hinted at by the formulas above).
In the end, you can reduce it down to an algebraic expression involving only $\sin \theta$ and $\cos \theta$, and since cosine is positive on the range of $\sin^{-1}$ and $\tan^{-1}$, you can convert that to strictly $\sin \theta$. Solve the resulting eequation for $\sin \theta$ (I haven't worked it that far, so I don't know how nasty that will be to do), and then back out the values of $\theta$.
