Help with $\frac{\sin\theta+\cos\theta}{-\sin\theta+\cos\theta}=?$ Assume that $0<\theta<\frac{\pi}{4}$. If $\sin2\theta=\frac{1}{4}$, then $\frac{\sin\theta+\cos\theta}{-\sin\theta+\cos\theta}=?$
My approach with this problem is to rationalize the denominator in the fraction to get this:
$$
\frac{\sin2\theta+1}{\cos2\theta}=\frac{\frac{1}{4}+1}{\cos2\theta}
$$
After that i got stuck with $\cos2\theta$ any advice how to attack the problem?

Alternative approach:
$$\frac{1}{4} = \sin(2\theta) = 2\sin(\theta)\cos(\theta). \tag1 $$
Also, since $0 < \theta < (\pi/4)$, you have that
$$\cos(2\theta) = \sqrt{1 - \left(\frac{1}{4}\right)^2} = \frac{\sqrt{15}}{4}. \tag2 $$
Finally, you have that
$$\cos^2(\theta) - \sin^2(\theta) = \cos(2\theta) = \frac{\sqrt{15}}{4}. \tag3 $$

You have that:
$$\frac{\cos(\theta) + \sin(\theta)}{\cos(\theta) - \sin(\theta)} \times 
\frac{\cos(\theta) - \sin(\theta)}{\cos(\theta) - \sin(\theta)}$$
$$= ~\frac{\cos^2(\theta) - \sin^2(\theta)}{\cos^2(\theta) - 2\sin(\theta)\cos(\theta) + \sin^2(\theta)}. \tag4 $$
Using (1), (2), and (3), the expression in (4) equals
$$\frac{\frac{\sqrt{15}}{4}}{1 - \frac{1}{4}} = \frac{\sqrt{15}}{3}.$$
