Question

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

Answer 1

Hint: In order to solve this question, we require some knowledge on the concept of the principal value, that is, for ${{\operatorname{cosec}}^{-1}}x$, if $\theta$ is the principal value of ${{\operatorname{cosec}}^{-1}}x$, then $-\dfrac{\pi }{2}\le \theta \le \dfrac{\pi }{2}$ where $\theta \ne 0$. Now, here, we will convert (-1) of ${{\operatorname{cosec}}^{-1}}\left( -1 \right)$ in terms of $\operatorname{cosec}\theta$, then we will find the principal value of ${{\operatorname{cosec}}^{-1}}\left( -1 \right)$.
Complete step-by-step answer:
In this question, we have been asked to find the principal value of ${{\operatorname{cosec}}^{-1}}\left( -1 \right)$. For that, we should have some idea about the cosecant ratios like $\operatorname{cosec}{{90}^{\circ }}=1$. Now, we know that $\operatorname{cosec}\left( -\theta \right)=-\operatorname{cosec}\left( \theta \right)$. So, we can express $-\operatorname{cosec}\left( {{90}^{\circ }} \right)$ as $\operatorname{cosec}\left( -{{90}^{\circ }} \right)$. And therefore, we can write, $-1=\operatorname{cosec}\left( -{{90}^{\circ }} \right)$. And we can further write it as, ${{\operatorname{cosec}}^{-1}}(-1)=\left( -{{90}^{\circ }} \right)$ that is in terms of inverse cosecant ratio. So, we can say that the principal value of ${{\operatorname{cosec}}^{-1}}\left( -1 \right)$ is $\left( -{{90}^{\circ }} \right)$. Now, we know that the range of principal value of ${{\operatorname{cosec}}^{-1}}x$ is \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]-\left\\{ 0 \right\\}. And we know that general solution of $\theta =\cos e{{c}^{-1}}x$ is $\theta =n\left( 180 \right)+{{\left( -1 \right)}^{n}}\cos e{{c}^{-1}}x$. Therefore, we can say that general solution of $\theta =n\left( 180 \right)-{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{2} \right)$. So, to get the principal, we will put a value of n as 0. Therefore, we get $\theta =-90$. And we know that ${{1}^{\circ }}=\dfrac{\pi }{180}$ radian. So we can say that ${{90}^{\circ }}=\dfrac{\pi }{2}$ radian. Hence, we can say that range of principal value of ${{\operatorname{cosec}}^{-1}}x$ is \left[ -{{90}^{\circ }},{{90}^{\circ }} \right]-\left\\{ 0 \right\\}. And we get the principal value as $\left( -{{90}^{\circ }} \right)$ which lies in the range of \left[ -{{90}^{\circ }},{{90}^{\circ }} \right]-\left\\{ 0 \right\\}.
Hence, we can say that the principal value of ${{\operatorname{cosec}}^{-1}}\left( -1 \right)$ as $\left( -{{90}^{\circ }} \right)$.

Note: In this question, the possible mistakes that the students can make is by writing $-1=\operatorname{cosec}\left( \dfrac{3\pi }{2} \right)$ which is actually not wrong but, $\dfrac{3\pi }{2}$ does not lie in the range of the principal value of ${{\operatorname{cosec}}^{-1}}x$ and so we have to convert that later in $\left( -\dfrac{\pi }{2} \right)$. So, it would be better to do it in the way used in the solution in the beginning itself.