Question

Find the principal value of ${{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$

Answer 1

Hint:First of all, take cosec on both sides and use $\operatorname{cosec}\left( {{\operatorname{cosec}}^{-1}}x \right)=x$. Now from the trigonometric table, find the value of $\theta$ at which $\operatorname{cosec}\theta =\dfrac{\sqrt{3}}{2}$ and from this find the principal value of the given expression.

Complete step-by-step answer:

Here, we have to find the principal value of ${{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$. Let us take the principal value of ${{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$ as y. So, we get,

$y={{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$

Now by taking cosec on both sides of the above equation, we get,

$\operatorname{cosec}y=\operatorname{cosec}\left( {{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right) \right)$

We know that $\operatorname{cosec}\left( {{\operatorname{cosec}}^{-1}}x \right)=x$. By using this in the RHS of the above equation, we get,

$\operatorname{cosec}y=\dfrac{2}{\sqrt{3}}....\left( i \right)$

Now, let us find the value at which $\operatorname{cosec}\theta =\dfrac{2}{\sqrt{3}}$ from the table below.

From the above table, we can see that

$\sin\dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$ and we know that $\sin\theta=\dfrac{1}{\operatorname{cosec}\theta}$

So, $\operatorname{cosec}\dfrac{\pi }{3}=\dfrac{2}{\sqrt{3}}$. So by substituting the value of $\dfrac{2}{\sqrt{3}}$ in equation (i), we get

$\operatorname{cosec}y=\operatorname{cosec}\dfrac{\pi }{3}$

We know that the range of values of ${{\operatorname{cosec}}^{-1}}\left( x \right)$ is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]-0$. So, we get,

$y=\dfrac{\pi }{3}$

Hence, we get the principal value of ${{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$ as $\dfrac{\pi }{3}$.

Note: In this question, students must ensure that the final value of y should be in the range of ${{\operatorname{cosec}}^{-1}}x$. Also, students can cross-check their answers by substituting the value of y in the initial equation as follows:

$y={{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$

We know that $y=\dfrac{\pi }{3}$, so we get,

$\dfrac{\pi }{3}={{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$

By taking cosec on both the sides, we get,

$\operatorname{cosec}\dfrac{\pi }{3}=\operatorname{cosec}\left( {{\operatorname{cosec}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right) \right)$

From the table, we know that $\operatorname{cosec}\dfrac{\pi }{3}=\dfrac{2}{\sqrt{3}}$. Also, we know that $\operatorname{cosec}\left( {{\operatorname{cosec}}^{-1}}\left( x \right) \right)=x$. By using this in the above equation, we get,

$\dfrac{2}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}$

Here, LHS = RHS. So, our answer is correct.