For the principal value, evaluate the following
\[{{\sin }^{-1}}\left[ \cos \left( 2{{\operatorname{cosec}}^{-1}}\left( -2 \right) \right) \right]\]

Hint:First of all, use \[{{\operatorname{cosec}}^{-1}}\left( -x \right)=-{{\operatorname{cosec}}^{-1}}x\] and then use a trigonometric table to find the value of \[{{\operatorname{cosec}}^{-1}}\left( 2 \right)\]. Now use \[\cos \left( -\theta \right)=\cos \theta \] and again use the table to find the value of \[\cos \dfrac{\pi }{3}\]. Find the angle at which \[\sin \theta =\dfrac{1}{2}\] from the table or the value of \[{{\sin }^{-1}}\left( \dfrac{1}{2} \right)\] to get the required answer.

Complete step-by-step answer:
In this question, we have to find the principal value of \[{{\sin }^{-1}}\left[ \cos \left( 2{{\operatorname{cosec}}^{-1}}\left( -2 \right) \right) \right]\].
First of all, let us consider the expression given in the question,
\[E={{\sin }^{-1}}\left[ \cos \left( 2{{\operatorname{cosec}}^{-1}}\left( -2 \right) \right) \right]\]
We know that, \[{{\operatorname{cosec}}^{-1}}\left( -x \right)=-{{\operatorname{cosec}}^{-1}}x\]. By using this in the above expression, we get,
\[E={{\sin }^{-1}}\left[ \cos \left( -2{{\operatorname{cosec}}^{-1}}2 \right) \right]....\left( i \right)\]
Now, let us draw the table for trigonometric ratios of general angles.

From the above table, we can see that,
\[\operatorname{cosec}\left( \dfrac{\pi }{6} \right)=2\]
\[\Rightarrow {{\operatorname{cosec}}^{-1}}\left( 2 \right)=\dfrac{\pi }{6}\]
So, by substituting the value of \[{{\operatorname{cosec}}^{-1}}\left( 2 \right)\] in the expression (i), we get,
\[E={{\sin }^{-1}}\left[ \cos \left( -2.\dfrac{\pi }{6} \right) \right]\]
\[E={{\sin }^{-1}}\left[ \cos \left( -\dfrac{\pi }{3} \right) \right]\]
We know that, \[\cos \left( -\theta \right)=\cos \theta \]. By using this in the above expression, we get,
\[E={{\sin }^{-1}}\left[ \cos \dfrac{\pi }{3} \right]\]
Now, from the trigonometric table, we can see that, \[\cos \dfrac{\pi }{3}=\dfrac{1}{2}\]. So, by substituting the value of \[\cos \dfrac{\pi }{3}\] in the above expression, we get,
\[E={{\sin }^{-1}}\left( \dfrac{1}{2} \right)....\left( ii \right)\]
Now we know that the range of principal value of \[{{\sin }^{-1}}\left( x \right)\] lies between \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\].
From the table of general trigonometric ratios, we get,
\[\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}\]
\[\Rightarrow {{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{6}\]
Now by substituting the value of \[{{\sin }^{-1}}\left( \dfrac{1}{2} \right)\] in the expression (ii), we get,
\[E=\dfrac{\pi }{6}\]
Hence, we get the value of \[{{\sin }^{-1}}\left[ \cos \left( 2{{\operatorname{cosec}}^{-1}}\left( -2 \right) \right) \right]\] as \[\dfrac{\pi }{6}\].

Note: In this question, students must take care that the value of the angle must lie in the range of \[{{\operatorname{cosec}}^{-1}}x\] which is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]-\left\{ 0 \right\}\] and \[{{\sin }^{-1}}x\] which is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\] accordingly. Also, students can verify their answer by equating the given expression with \[\dfrac{\pi }{6}\] and taking sin on both sides and keep solving until LHS = RHS.