If $$0 < \theta < \frac{\pi}{2}$$ and $$\tan \theta + \sec \theta = 3$$, then $$\sin \theta =$$

We have $$\tan\ \theta\ +\sec\theta\ =3$$

$$\frac{\sin\theta}{\cos\theta\ }\ \ +\frac{1}{\cos\theta\ }=3\ \ or\ 1+\sin\theta\ =3\cos\theta\ $$

Squaring both sides

$$1+\sin^2\theta\ +2\sin\theta\ =9\cos^2\theta\ $$

$$1+\sin^2\theta\ +2\sin\theta\ =9-9\sin^2\theta\ \ $$

$$10\sin^2\theta\ +2\sin\theta\ -8=0\ or\ 5\ \sin^2\theta\ +\sin\theta\ -4=0$$

$$\ \ 5\ \sin^2\theta\ +5\sin\theta\ -4\sin\theta\ \ -4=0$$

$$5\sin\theta\ \left(\sin\theta\ +1\right)-4\left(\sin\theta\ +1\right)=0$$

$$\left(5\sin\theta\ -4\right)\left(\sin\theta\ +1\right)=0$$

$$0<\theta\ <\pi\ \ so\ \sin\theta\ \ne\ -1\ $$

Hence $$\sin\theta\ =\frac{4}{5}$$