Solving without squaring

Question

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 $\sin 2\theta = \dfrac{-3}{2}\dfrac{(1-\tan\theta/2)^2}{1+\tan^2(\theta/2)}$

Answer 1

$$\implies3(\sin\theta-\cos\theta)=4\sin\theta\cos\theta=2\sin2\theta$$

$$\iff3\sqrt2\sin\left(\theta-\dfrac\pi4\right)=2\cos2\left(\theta-\dfrac\pi4\right)=2\left(1-2\sin^2\left(\theta-\dfrac\pi4\right)\right)$$

$$\iff4\sin^2\left(\theta-\dfrac\pi4\right)+3\sqrt2\sin\left(\theta-\dfrac\pi4\right)-2=0$$

$$\sin\left(\theta-\dfrac\pi4\right)=\dfrac{-3\sqrt2\pm5\sqrt2}8=-\sqrt2\text{ or } \dfrac1{2\sqrt2}$$

As $\sin\left(\theta-\dfrac\pi4\right)\ge-1,$

$$\sin\left(\theta-\dfrac\pi4\right)=\dfrac1{2\sqrt2}$$

$$\implies\sin2\theta=\cos2\left(\theta-\dfrac\pi4\right)=1-2\sin^2\left(\theta-\dfrac\pi4\right)=?$$

Answer 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!