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
$\sin^2\theta=$

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.
