Question

For the principal value, evaluate the following
\[{{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)+{{\operatorname{cosec}}^{-1}}\left( -\dfrac{2}{\sqrt{3}} \right)\].

Answer 1

Hint:First of all, use \[{{\operatorname{cosec}}^{-1}}\left( -x \right)=-{{\operatorname{cosec}}^{-1}}x\] and \[{{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x\] to simplify the given expression. Now from the trigonometric table, find the value of \[\theta \] at which \[\sin \theta =\dfrac{\sqrt{3}}{2}\] and \[\operatorname{cosec}\theta =\dfrac{2}{\sqrt{3}}\] or the value of \[{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\] and \[{{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)\] and substitute these in the given expression to get the required value.

Complete step-by-step answer:
In this question, we have to find the principal value of \[{{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)+{{\operatorname{cosec}}^{-1}}\left( -\dfrac{2}{\sqrt{3}} \right)\].
First of all, let us consider the expression given in the question,
\[E={{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)+{{\operatorname{cosec}}^{-1}}\left( -\dfrac{2}{\sqrt{3}} \right)\]
We know that, \[{{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x\]. By using this in the above expression, we get,
\[E=-{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)+{{\operatorname{cosec}}^{-1}}\left( -\dfrac{2}{\sqrt{3}} \right)\]
We know that, \[{{\operatorname{cosec}}^{-1}}\left( -x \right)=-{{\operatorname{cosec}}^{-1}}x\]. By using this in the above expression, we get,
\[E=-{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)-{{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right).....\left( i \right)\]
Now, let us draw the table for trigonometric ratios of general angles.

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,
\[\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}\]
By taking \[{{\sin }^{-1}}\] on both the sides, we get,
\[{{\sin }^{-1}}\sin \left( \dfrac{\pi }{3} \right)={{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\]
We know that for \[\dfrac{-\pi }{2}\le x\le \dfrac{\pi }{2},{{\sin }^{-1}}\sin \left( x \right)=x\]. So, we get,
\[\dfrac{\pi }{3}={{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)...\left( ii \right)\]
Now, we also know that the range of principal value of \[{{\operatorname{cosec}}^{-1}}x\] lies between \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]-\left\{ 0 \right\}\]
From the table of trigonometric ratios, we get,
\[\operatorname{cosec}\left( \dfrac{\pi }{3} \right)=\dfrac{2}{\sqrt{3}}\]
By taking \[{{\operatorname{cosec}}^{-1}}\] on both the sides, we get,
\[{{\operatorname{cosec}}^{-1}}\left( \operatorname{cosec}\left( \dfrac{\pi }{3} \right) \right)={{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)\]
We know that for \[\left[ \dfrac{-\pi }{2}\le x\le \dfrac{\pi }{2} \right]-\left\{ 0 \right\},{{\operatorname{cosec}}^{-1}}\operatorname{cosec}\left( x \right)=x\]. So, we get,
\[\dfrac{\pi }{3}={{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)....\left( iii \right)\]
So, by substituting the value of \[{{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)\] from equation (iii) and \[{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\] from equation (ii) in equation (i), we get,
\[E=-\dfrac{\pi }{3}-\dfrac{\pi }{3}\]
\[E=\dfrac{-2\pi }{3}\]
Hence, we get the value of \[{{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)+{{\operatorname{cosec}}^{-1}}\left( -\dfrac{2}{\sqrt{3}} \right)\] as \[\dfrac{-2\pi }{3}\].

Note: In this question, many students make this mistake of taking \[{{\operatorname{cosec}}^{-1}}\left( -x \right)\] as \[\pi -{{\operatorname{cosec}}^{-1}}\left( x \right)\] like that in case of \[{{\sec }^{-1}}\left( -x \right)\] and \[{{\cot }^{-1}}\left( -x \right)\] which is wrong because \[{{\operatorname{cosec}}^{-1}}\left( -x \right)=-{{\operatorname{cosec}}^{-1}}\left( x \right)\].Students must strictly take care of the domain and range of the inverse trigonometric functions. Also, students must take care that the angle they take must lie in the range of the respective trigonometric functions to get the required answer.