Question

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

Answer 1

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 from the trigonometric ratio table, check the value of $\operatorname{cosec}\dfrac{\pi }{4}$ 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 ${{\operatorname{cosec}}^{-1}}\left( -\sqrt{2} \right)$. Let us consider the value of ${{\operatorname{cosec}}^{-1}}\left( -\sqrt{2} \right)$ as y. So, we get,
$y={{\operatorname{cosec}}^{-1}}\left( -\sqrt{2} \right)$
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,
$y=-{{\operatorname{cosec}}^{-1}}\left( \sqrt{2} \right)$
Now by taking cosec on both sides of the above equation, we get,
$\operatorname{cosec}y=\operatorname{cosec}\left( -{{\operatorname{cosec}}^{-1}}\sqrt{2} \right)$
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,
$\operatorname{cosec}y=-\operatorname{cosec}\left( {{\operatorname{cosec}}^{-1}}\sqrt{2} \right)$
We know that $\operatorname{cosec}\left( {{\operatorname{cosec}}^{-1}}x \right)=x$. By using this in the RHS of the above equation, we get,
$\operatorname{cosec}y=-\sqrt{2}....\left( i \right)$

From the above table, we can see that,
$\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}$ and we know that $\sin\theta=\dfrac{1}{\operatorname{cosec}\theta}$
So, $\operatorname{cosec}\dfrac{\pi }{4}=\sqrt{2}....\left( ii \right)$
By multiplying – 1 on both sides of equation (ii), we get
$-\operatorname{cosec}\dfrac{\pi }{4}=-\sqrt{2}$
We know that $\operatorname{cosec}\left( -x \right)=-\operatorname{cosec}x$. By using this in the above equation, we can write,
$\operatorname{cosec}\left( \dfrac{-\pi }{4} \right)=-\sqrt{2}$
Now, by substituting the value of $-\sqrt{2}$ in terms of cosec in the above equation, we get,
$\operatorname{cosec}y=\operatorname{cosec}\left( \dfrac{-\pi }{4} \right)$
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,
$y=\dfrac{-\pi }{4}$
Hence, we get the principal value of ${{\operatorname{cosec}}^{-1}}\left( -\sqrt{2} \right)$ as $\dfrac{-\pi }{4}$.

Note: Students must remember the range and domain of inverse trigonometric functions. Also students must remember the values of $\sin \theta ,\cos \theta$, etc. at general angles like ${{0}^{o}},\dfrac{\pi }{4},\dfrac{\pi }{3}$, etc. Some students make this mistake of writing ${{\operatorname{cosec}}^{-1}}\left( -x \right)$ as $\pi -{{\operatorname{cosec}}^{-1}}\left( x \right)$ which is wrong because ${{\operatorname{cosec}}^{-1}}\left( -x \right)=-{{\operatorname{cosec}}^{-1}}\left( x \right)$. So this must be taken care of.