Given that \(0\le\theta<\dfrac\pi2\), show that \(\dfrac{\sin\theta}{\sqrt{\operatorname{cosec}^2\theta-1}}+\dfrac1{\sqrt{1+\tan^2\theta}}\) can be written as \(\sec\theta\).[4]
Given that \(\sec x=\alpha\), where \(\dfrac{3\pi}2<x\le2\pi\), find \(\sin x\) in terms of \(\alpha\).[3]
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Answer:(a) \(\sec\theta\), where the original expression is defined (shown) (b) \(-\sqrt{1-\dfrac1{\alpha^2}}\)