Evaluate the following:
(i) \[\sin ({{\cot }^{-1}}x)\]
(ii) \[\sin \left( \dfrac{\pi }{2}-{{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right) \right)\]

Hint: In the first part we will assume \[{{\cot }^{-1}}x=\theta \] and then we will convert it in terms of sin and cos and use the formula \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to get the answer. Then in the second part of the question we will use \[{{\sin }^{-1}}(-x)=-{{\sin }^{-1}}x\] and then substitute the angle for which the value is given and get our answer.
Complete step-by-step answer:
(i) The expression mentioned in the first part of the question is \[\sin ({{\cot }^{-1}}x).........(1)\]
Now let \[{{\cot }^{-1}}x=\theta ........(2)\]
Now rearranging equation (2) we get,
\[\cot \theta =x........(3)\]
Now converting cot in terms of cos and sin in equation (3) we get,
\[\dfrac{\cos \theta }{\sin \theta }=x........(4)\]
Now squaring both sides in equation (4) we get,
\[\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }={{x}^{2}}........(5)\]
Now cross multiplying the terms in equation (5) we get,
\[{{\cos }^{2}}\theta ={{x}^{2}}{{\sin }^{2}}\theta ........(6)\]
Now we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and hence using this in equation (6) we get,
\[1-{{\sin }^{2}}\theta ={{x}^{2}}{{\sin }^{2}}\theta ........(7)\]
Now simplifying and rearranging the terms in equation (7) and solving we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}{{\sin }^{2}}\theta +{{\sin }^{2}}\theta =1 \\
 & \Rightarrow {{\sin }^{2}}\theta (1+{{x}^{2}})=1 \\
 & \Rightarrow {{\sin }^{2}}\theta =\dfrac{1}{1+{{x}^{2}}}........(8) \\
\end{align}\]
Taking square root of both sides of equation (8) we get,
\[\Rightarrow \sin \theta =\pm \dfrac{1}{\sqrt{1+{{x}^{2}}}}........(9)\]
Now substituting from equation (2) in equation (9) we get,
\[\Rightarrow \sin ({{\cot }^{-1}}x)=\pm \dfrac{1}{\sqrt{1+{{x}^{2}}}}\]
Hence \[\pm \dfrac{1}{\sqrt{1+{{x}^{2}}}}\] is the answer.
(ii) The expression mentioned in the second part of the question is \[\sin \left( \dfrac{\pi }{2}-{{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right) \right)\].
\[\Rightarrow \sin \left( \dfrac{\pi }{2}-{{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right) \right).........(1)\]
We know that the domain of \[{{\sin }^{-1}}x\] is between -1 and +1. Also the range is between \[-\dfrac{\pi }{2}\] and \[\dfrac{\pi }{2}\].
Also we know that \[{{\sin }^{-1}}(-x)=-{{\sin }^{-1}}x\] and using this in equation (1) we get,
\[\Rightarrow \sin \left( \dfrac{\pi }{2}+{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right).........(2)\]
We know that \[\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}\] and substituting this in equation (2) we get,
\[\Rightarrow \sin \left( \dfrac{\pi }{2}+{{\sin }^{-1}}\left( \sin \dfrac{\pi }{3} \right) \right).........(3)\]
Now simplifying and rearranging in equation (3) we get,
\[\Rightarrow \sin \left( \dfrac{\pi }{2}+\dfrac{\pi }{3} \right).........(4)\]
Now taking the LCM and solving in equation (4) we get,
\[\Rightarrow \sin \left( \dfrac{5\pi }{6} \right)=\sin \left( \pi -\dfrac{\pi }{6} \right)=\sin \dfrac{\pi }{6}=\dfrac{1}{2}\]
Hence \[\dfrac{1}{2}\] is the answer.

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We in a hurry can make a mistake in applying the cofunction identity \[\sin \left( \pi -x \right)=\sin x\] as we can write cos in place of sin in equation (4) in part (ii).