Question

How do you find the exact value of $\tan \left( {{\sin }^{-1}}\left( \dfrac{2}{3} \right) \right)$?

Answer 1

We will look at the definition of inverse trigonometric function. We will use this to get an expression that gives us the value of a trigonometric function, here, it is the sine function. Then, we will use the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to find the value of the cosine function using the sine function. Then we will use these values to find the value of the tangent function.

Complete answer:
The inverse trigonometric function is defined as the inverse function of a trigonometric function and it is used to find the value of the angle for a given trigonometric ratio. Now, we will take the inverse sine function inside the bracket first.
Let $x={{\sin }^{-1}}\left( \dfrac{2}{3} \right)$. So, we have $\sin x=\dfrac{2}{3}$. We have the identity with the sine and cosine functions given as ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Using this identity, we will find the value of the cosine function for the angle $x$. Let us substitute the value $\sin x=\dfrac{2}{3}$ in this identity. So, we have the following,
${{\left( \dfrac{2}{3} \right)}^{2}}+{{\cos }^{2}}x=1$
Simplifying this equation, we get
$\begin{aligned} & \dfrac{4}{9}+{{\cos }^{2}}x=1 \\\ & \Rightarrow {{\cos }^{2}}x=1-\dfrac{4}{9} \\\ & \therefore {{\cos }^{2}}x=\dfrac{5}{9} \\\ \end{aligned}$
Taking the square root on both sides, we have
$\Rightarrow \cos x=\dfrac{\sqrt{5}}{3}$
Let us look at the definition of the tangent function. The tangent function is defined as
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Since $x={{\sin }^{-1}}\left( \dfrac{2}{3} \right)$, we have that $\tan \left( {{\sin }^{-1}}\left( \dfrac{2}{3} \right) \right)=\tan x$. Using the definition of the tangent function, we get the following,
$\Rightarrow \tan \left( {{\sin }^{-1}}\left( \dfrac{2}{3} \right) \right)=\dfrac{\sin x}{\cos x}$
Substituting the values of the sine and cosine functions, we get
$\begin{aligned} & \Rightarrow \tan \left( {{\sin }^{-1}}\left( \dfrac{2}{3} \right) \right)=\dfrac{\left( \dfrac{2}{3} \right)}{\left( \dfrac{\sqrt{5}}{3} \right)} \\\ & \therefore \tan \left( {{\sin }^{-1}}\left( \dfrac{2}{3} \right) \right)=\dfrac{2}{\sqrt{5}} \\\ \end{aligned}$

Note: It is important to be familiar with the definitions and relations of the trigonometric functions. There are multiple identities that give relations between different trigonometric functions. These identities are useful in solving or simplifying equations containing trigonometric functions. We should do the calculations explicitly to avoid making minor mistakes.