Answer: \(\frac{\sin\theta}{\sqrt{\operatorname{cosec}^2\theta-1}}+\frac{1}{\sqrt{1+\tan^2\theta}}=\sec\theta\), and \(\sin x=-\sqrt{1-\frac{1}{\alpha^2}}\).

Since \(0\leqslant\theta\lt \frac{\pi}{2}\), the relevant trigonometric ratios are non-negative.

Use

\(\operatorname{cosec}^2\theta-1=\cot^2\theta,\quad 1+\tan^2\theta=\sec^2\theta.\)

Then

\(\frac{\sin\theta}{\sqrt{\operatorname{cosec}^2\theta-1}}+\frac{1}{\sqrt{1+\tan^2\theta}} =\frac{\sin\theta}{\cot\theta}+\frac{1}{\sec\theta}.\)

This becomes

\(\sin\theta\tan\theta+\cos\theta.\)

Now

\(\sin\theta\tan\theta+\cos\theta =\frac{\sin^2\theta}{\cos\theta}+\cos\theta =\frac{\sin^2\theta+\cos^2\theta}{\cos\theta} =\frac{1}{\cos\theta} =\sec\theta.\)

For the second part, \(\sec x=\alpha\), so

\(\cos x=\frac{1}{\alpha}.\)

Using \(\sin^2x+\cos^2x=1\),

\(\sin^2x=1-\frac{1}{\alpha^2}.\)

Since \(\frac{3\pi}{2}\lt x\leqslant2\pi\), \(x\) is in the fourth quadrant, so \(\sin x\leqslant0\). Therefore

\(\sin x=-\sqrt{1-\frac{1}{\alpha^2}}.\)