Question

Find the principal value of ${{\operatorname{cosec}}^{-1}}\left( -2 \right)$

Answer 1

Hint: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.