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

WLOG let $\theta=\dfrac\pi2-2y\implies x=\csc2y+\cot2y=\dfrac{1+\cos2y}{\sin2y}=\cot y$
$$\sin\theta=\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}=\dfrac{\cot^2y-1}{\cot^2y+1}=?$$
