If $\sec \theta+\tan \theta= \sqrt{3}$ then the positive value of $\sin \theta$ If $\sec \theta+\tan \theta=\sqrt{3}$ then the positive value of $\sin \theta$
Note:
$1/\cos\theta+\sin\theta/\cos \theta=\sqrt{3}$
$\sin\theta=\sqrt{3}\cos \theta-1$
squaring on both sides we get
$\sin^2\theta=$
 A: Since $1 = \sec^2 \theta - \tan^2 \theta = (\sec \theta + \tan \theta)(\sec \theta - \tan \theta)$, we have $\sec \theta - \tan \theta = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$. 
Now, we have two equations: 
(1) $\sec \theta + \tan \theta = \sqrt{3}$
(2) $\sec \theta - \tan \theta = \dfrac{\sqrt{3}}{3}$
Adding the two gives $\sec \theta = \dfrac{2\sqrt{3}}{3}$. Subtracting the 2nd from the 1st gives $\tan \theta = \dfrac{\sqrt{3}}{3}$. 
Can you find $\sin \theta$ from this? Note that this method doesn't yield any extraneous solutions.
A: Better: write it as
$$\sin\theta+1=\sqrt3\,\cos\theta$$
and then square to obtain a quadratic in $\sin\theta$.  Can you finish it from here?
A: Beginning with $1+\sin \theta = \sqrt{3} \cos \theta$, we can use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to obtain $$1+\sin \theta = \sqrt{3(1-\sin^2\theta)} \\ \Rightarrow 1 + 2\sin \theta + \sin^2 \theta = 3-3 \sin^2 \theta \\ \Rightarrow 4\sin^2 \theta + 2 \sin \theta - 2 = 0$$ Now you have an quadratic equation for $\sin \theta$ which you can solve by factoring so: $$4\sin^2 \theta + 2 \sin \theta - 2 = 0 \\ \Rightarrow 4(\sin \theta + 1)(\sin \theta - \frac{1}{2})=0 \\ \Rightarrow \sin \theta = -1, \sin \theta = \frac{1}{2}$$ Since we want the positive value, we conclude that $\sin \theta = 1/2$
