Find the principal value of \[{{\operatorname{cosec}}^{-1}}\left( -2 \right)\]
VerifiedHint:First of all, use \[{{\operatorname{cosec}}^{-1}}\left( -x \right)=-{{\operatorname{cosec}}^{-1}}x\]. Now, take cosec on both sides and use \[\operatorname{cosec}\left( -\theta \right)=-\operatorname{cosec}\theta \]. Now use the value of \[\operatorname{cosec}\dfrac{\pi }{6}\] from the trigonometric ratio table and get the principal value of the given expression in the range of \[{{\operatorname{cosec}}^{-1}}x\].
Complete step-by-step answer:
Here, we have to find the principal value of \[{{\operatorname{cosec}}^{-1}}\left( -2 \right)\]. Let us consider the value of \[{{\operatorname{cosec}}^{-1}}\left( -2 \right)\] as y. So, we get,
\[y={{\operatorname{cosec}}^{-1}}\left( -2 \right)\]
We know that \[{{\operatorname{cosec}}^{-1}}\left( -x \right)=-{{\operatorname{cosec}}^{-1}}x\]. By using this in the RHS of the above equation, we get,
\[y=-{{\operatorname{cosec}}^{-1}}\left( 2 \right)\]
Now by taking cosec on both sides of the above equation, we get,
\[\operatorname{cosec}y=\operatorname{cosec}\left( -{{\operatorname{cosec}}^{-1}}2 \right)\]
We know that \[{{\operatorname{cosec}}^{-1}}\left( -x \right)=-{{\operatorname{cosec}}^{-1}}x\]. By using this in the RHS of the above equation, we get,
\[\operatorname{cosec}y=-\operatorname{cosec}\left( {{\operatorname{cosec}}^{-1}}2 \right)\]
We know that \[\operatorname{cosec}\left( {{\operatorname{cosec}}^{-1}}x \right)=x\]. By using this in the RHS of the above equation, we get,
\[\operatorname{cosec}y=-2....\left( i \right)\]
From the above table, we can see that,
$\sin\dfrac{\pi}{6}=\dfrac{1}{2}$ and we know that $\sin\theta=\dfrac{1}{\operatorname{cosec}\theta}$
So ,\[\operatorname{cosec}\dfrac{\pi }{6}=2....\left( ii \right)\]
By multiplying – 1 on both sides of equation (ii), we get
\[-\operatorname{cosec}\dfrac{\pi }{6}=-2\]
We know that \[\operatorname{cosec}\left( -x \right)=-\operatorname{cosec}x\]. By using this in the above equation, we can write,
\[\operatorname{cosec}\left( \dfrac{-\pi }{6} \right)=-2\]
Now, by substituting the value of - 2 in terms of cosec in the above equation, we get,
\[\operatorname{cosec}y=\operatorname{cosec}\left( \dfrac{-\pi }{6} \right)\]
We know that the range of principal values of \[{{\operatorname{cosec}}^{-1}}\left( x \right)\] is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]-0\]. So, we get,
\[y=\dfrac{-\pi }{6}\]
Hence, we get the principal value of \[{{\operatorname{cosec}}^{-1}}\left( -2 \right)\] as \[\dfrac{-\pi }{6}\].
Note: In this question, students can cross-check their answer by substituting the value of y in the initial equation and then checking if LHS = RHS after taking cosec on both sides. Also, students must remember the range of all the inverse trigonometric functions. Also, students must remember the values of various trigonometric ratios like \[\sin \theta ,\cos \theta \], etc. and get general angles like \[{{0}^{o}},{{30}^{o}},{{60}^{o}}\], etc.
