Solving without squaring If $\sec \theta - \csc \theta = \dfrac{4}{3}$, then prove that $\theta= \dfrac{1}{2}\arcsin {\dfrac{3}{4}}$
I have tried really hard to solve this without squaring both sides of the equation but it seems next to impossible. 
The closest thing I reached is $\sin \theta = \dfrac{3(1- 2\sin^2(\theta/2))}{8\sin^2(\theta/2)- 1}$
I also obtained $\sin 2\theta = \dfrac{-3}{2}\dfrac{(1-\tan\theta/2)^2}{1+\tan^2(\theta/2)}$

Let $\sin{\theta}-\cos{\theta}=t$.
Thus, $\sin{\theta}\cos{\theta}=\frac{1-t^2}{2}$ and we have
$$\frac{2t}{1-t^2}=\frac{4}{3}$$ or
$$2t^2+3t-2=0$$ and since by C-S $$|\sin{\theta}-\cos{\theta}|\leq\sqrt{(1^2+1^2)(\sin^2\theta+\cos^2\theta)}=\sqrt2,$$ we obtain $t=\frac{1}{2}.$
Thus, $$\sin2\theta=2\cdot\frac{1-\frac{1}{4}}{2}=\frac{3}{4}$$ and we are done!
