If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$. If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.
$$\sec \theta + \tan \theta = x$$
$$\dfrac {1}{\cos \theta }+\dfrac {\sin \theta }{\cos \theta }=x$$
$$\dfrac {1+\sin \theta }{\sqrt {1-\sin^2 \theta }}=x$$
$$1+\sin \theta =x\sqrt {1-\sin^2 \theta }$$
$$1+2\sin \theta + \sin^2 \theta = x^2-x^2 \sin^2 \theta $$
$$x^2 \sin^2 \theta + \sin^2 \theta + 2\sin \theta = x^2-1$$
$$\sin^2 \theta (x^2+1) + 2\sin \theta =x^2-1$$

Here is a different approach:  Since $1 + \tan^2\theta = \sec^2\theta$, we have 
$$\sec^2\theta - \tan^2\theta = 1$$
Factoring yields
$$(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$$
Since we are given that $\sec\theta + \tan\theta = x$, we obtain
$$x(\sec\theta - \tan\theta) = 1$$
Therefore, 
$$\sec\theta - \tan\theta = \frac{1}{x}$$
This yields the system of equations
\begin{align*}
\sec\theta + \tan\theta & = x \tag{1}\\
\sec\theta - \tan\theta & = \frac{1}{x} \tag{2}
\end{align*}
Adding equations 1 and 2 and solving for $\sec\theta$ yields
\begin{align*}
2\sec\theta & = x + \frac{1}{x}\\
2\sec\theta & = \frac{x^2 + 1}{x}\\
\sec\theta & = \frac{x^2 + 1}{2x} 
\end{align*}
Therefore, 
$$\cos\theta = \frac{1}{\sec\theta} = \frac{2x}{x^2 + 1}$$
Subtracting equation 2 from equation 1 and solving for $\tan\theta$ yields
\begin{align*}
2\tan\theta & = x - \frac{1}{x}\\
2\tan\theta & = \frac{x^2 - 1}{x}\\
\tan\theta & = \frac{x^2 - 1}{2x}
\end{align*}
Thus,
$$\sin\theta = \tan\theta\cos\theta = \frac{x^2 - 1}{2x} \cdot \frac{2x}{x^2 + 1} = \frac{x^2 - 1}{x^2 + 1}$$
