If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$. If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.
$$\sec \theta + \tan \theta = x$$
$$\dfrac {1}{\cos \theta }+\dfrac {\sin \theta }{\cos \theta }=x$$
$$\dfrac {1+\sin \theta }{\sqrt {1-\sin^2 \theta }}=x$$
$$1+\sin \theta =x\sqrt {1-\sin^2 \theta }$$
$$1+2\sin \theta + \sin^2 \theta = x^2-x^2 \sin^2 \theta $$
$$x^2 \sin^2 \theta + \sin^2 \theta + 2\sin \theta = x^2-1$$
$$\sin^2 \theta (x^2+1) + 2\sin \theta =x^2-1$$
 A: From where you are:
You obtained a quadratic function in $\sin(\theta)$. Perform the substitution $u =\sin(\theta)$. 
We obtain the quadratic (in $u$):
$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$ 
$$\Rightarrow u_{1,2} = \frac{- 2 \pm \sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$
$$ = \frac{- 2 \pm \sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$
$$ = \frac{- 2 \pm \sqrt{4x^4}}{2(x^2+1)}$$
$$ = \frac{- 2 \pm 2x^2}{2(x^2+1)}$$
$$ = \frac{- 1 \pm x^2}{x^2+1}$$
Therefore, 
$$\sin(\theta)_{1,2} = \frac{- 1 \pm x^2}{x^2+1}$$
One of those solutions will not work out.
This happened because you squared the equation multiple times and we know that $a = b$ is not equivalent with $a^2 = b^2$, so you should fill in both solutions in the original expression and see which one works and which one doesn't.
A: 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,
$$\sin\theta = \tan\theta\cos\theta = \frac{x^2 - 1}{2x} \cdot \frac{2x}{x^2 + 1} = \frac{x^2 - 1}{x^2 + 1}$$
A: 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
$$
X=\frac{1}{x}(1+Y)=\frac{1}{x}\frac{x^2+1+x^2-1}{x^2+1}=\frac{2x}{x^2+1}
$$
A: WLOG let $\theta=\dfrac\pi2-2y\implies x=\csc2y+\cot2y=\dfrac{1+\cos2y}{\sin2y}=\cot y$
$$\sin\theta=\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}=\dfrac{\cot^2y-1}{\cot^2y+1}=?$$
