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

The equation becomes
$$
1+\sin\theta=x\cos\theta
$$
Set $X=\cos\theta$ and $Y=\sin\theta$, so the equation becomes
$$
\begin{cases}
X^2+Y^2=1 \\[4px]
1+Y=xX
\end{cases}
$$
Note that $x\ne0$ and substitute $X=x^{-1}(1+Y)$ in the first equation getting
$$
(1+Y)^2+x^2Y^2=x^2
$$
that simplifies to
$$
(1+x^2)Y^2+2Y+1-x^2=0
$$
that yields
$$
Y=-1 \qquad\text{or}\qquad Y=\frac{x^2-1}{x^2+1}
$$
Is $Y=-1$ a solution for the problem?
By the way, you also get $\cos\theta$, since
$$
X=\frac{1}{x}(1+Y)=\frac{1}{x}\frac{x^2+1+x^2-1}{x^2+1}=\frac{2x}{x^2+1}
$$
