Question

How do you evaluate $\tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right)$ without a calculator?

Answer 1

Hint: This problem deals with applying the basic and important trigonometric identities. We are given a tangent trigonometric expression inside of which there is an inverse of cosine trigonometric expression of a particular value. So in order to proceed to get the exact value of the expression, first we need to assign the given inverse cosine trigonometric value to a variable, and then solve.

Complete step-by-step solution:
Given the expression of trigonometric and inverse trigonometric ratio which is $\tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right)$.
Consider the given expression, as given below:
$ \Rightarrow \tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right)$
Now consider the inside of tangent value which is ${\cos ^{ - 1}}\left( { - \dfrac{2}{3}} \right)$, as given below:
Let the expression of ${\cos ^{ - 1}}\left( { - \dfrac{2}{3}} \right)$ is equal to $\alpha $, which is mathematically expressed below:
$ \Rightarrow \alpha = {\cos ^{ - 1}}\left( { - \dfrac{2}{3}} \right)$
Now take inverse cosine trigonometric function on both the sides of the above equation, as shown below:
$ \Rightarrow \cos \alpha = \cos \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right)$
Here on the right hand side of the above equation, cosine and inverse cosine trigonometric function gets cancelled, as shown below:
$ \Rightarrow \cos \alpha = - \dfrac{2}{3}$
Now as we considered $\alpha = {\cos ^{ - 1}}\left( { - \dfrac{2}{3}} \right)$, hence the expression $\tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right)$ becomes as shown below:
\[ \Rightarrow \tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right) = \tan \alpha \]
So if we find the value of $\tan \alpha $, then it is the same as finding the value of$\tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right)$.
Hence finding the value of $\tan \alpha $.
But we know the value of $\cos \alpha $, which is equal to $ - \dfrac{2}{3}$.
Hence to find the value of $\tan \alpha $, we can express $\tan \alpha $in terms of $\cos \alpha $, and then can get the value of $\tan \alpha $.
So expressing $\tan \alpha $ in terms of $\cos \alpha $, as given below:
$ \Rightarrow \tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}$
We know that from the basic trigonometric identity ${\sin ^2}\alpha + {\cos ^2}\alpha = 1$, from here the value of $\sin \alpha $ can be written as:
$ \Rightarrow {\sin ^2}\alpha = 1 - {\cos ^2}\alpha $
$\therefore \sin \alpha = \sqrt {1 - {{\cos }^2}\alpha } $
Now substituting the above expression in the expression of $\tan \alpha $, as given below:
$ \Rightarrow \tan \alpha = \dfrac{{\sqrt {1 - {{\cos }^2}\alpha } }}{{\cos \alpha }}$
We obtained that the value of $\cos \alpha = - \dfrac{2}{3}$, hence substituting it in the above expression, as shown:
$ \Rightarrow \tan \alpha = \dfrac{{\sqrt {1 - {{\left( { - \dfrac{2}{3}} \right)}^2}} }}{{\left( { - \dfrac{2}{3}} \right)}}$
Simplifying the above expression, as given below:
$ \Rightarrow \tan \alpha = - \dfrac{3}{2}\sqrt {1 - \left( {\dfrac{4}{9}} \right)} $
$ \Rightarrow \tan \alpha = - \dfrac{3}{2}\sqrt {\dfrac{5}{9}} $
As we know that the value of under root of 9 is 3, $\sqrt 9 = 3$, as shown:
$ \Rightarrow \tan \alpha = - \dfrac{3}{{2 \times 3}}\sqrt 5 $
$ \Rightarrow \tan \alpha = - \dfrac{{\sqrt 5 }}{2}$

Hence the value of $\tan \left( {{{\cos }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right) = - \dfrac{{\sqrt 5 }}{2}$.

Note: Please note that while solving any trigonometric based problems, we need to be through with all the important and basic trigonometric identities, few are given below:
$ \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = 1$
From which we can obtain $\sin \alpha = \sqrt {1 - {{\cos }^2}\alpha } $
$ \Rightarrow {\sec ^2}\alpha - {\tan ^2}\alpha = 1$
From which we can obtain \[\tan \alpha = \sqrt {{{\sec }^2}\alpha - 1} \]
$ \Rightarrow \cos e{c^2}\alpha - {\cot ^2}\alpha = 1$
From which we can obtain $\cot \alpha = \sqrt {\cos e{c^2}\alpha - 1} $