Question

How do you evaluate $\tan \left( {\arccos \left( {\dfrac{2}{3}} \right)} \right)$?

Answer 1

Here in this question, we have to evaluate the given function and find the value, this can be solve by, firstly we have to take substitution for $\arccos \left( {\dfrac{2}{3}} \right)$ and later using the definition of the tan function and using the one of standard identity of trigonometry i.e., ${\sin ^2}\theta + {\cos ^2}\theta = 1$ on further simplification we get the required solution.

Complete step-by-step solution:
The question is related to the trigonometry and it includes the trigonometry ratios. The trigonometry ratios are sine, cosine, tangent, cosecant, secant and cotangent. These trigonometry ratios are abbreviated as sin, cos, tan, csc, sec and cot.
Now consider the given trigonometric function:
$\Rightarrow \,\,\,\tan \left( {\arccos \left( {\dfrac{2}{3}} \right)} \right)$------(1)
Now take substitution for
$\Rightarrow \alpha = \arccos \left( {\dfrac{2}{3}} \right)$ or $co{s^{ - 1}}\left( {\dfrac{2}{3}} \right) = \alpha$
The equation (1) becomes:
$\Rightarrow \,\,\,\tan \left( \alpha \right)$
As, we know the definition of the tangent function or tan i.e., tan function is the ratio of sine and cosine function i.e., $\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}$--------(2)
Now, we have to find the value of $\sin \alpha$ and $\cos \alpha$.
Consider,
$\Rightarrow \,\,\,co{s^{ - 1}}\left( {\dfrac{2}{3}} \right) = \alpha$
Take cos on both side, we get
$\Rightarrow \,\,\,cos\alpha = \dfrac{2}{3}$
As, we know the standard trigonometric identity: ${\sin ^2}\alpha + {\cos ^2}\alpha = 1$ or
$\Rightarrow \,\,\sin \alpha = \sqrt {1 - {{\cos }^2}\alpha }$
Substitute the value of $\cos \alpha$, then
$\Rightarrow \,\,\sin \theta = \sqrt {1 - {{\left( {\dfrac{2}{3}} \right)}^2}}$
$\Rightarrow \,\,\sin \theta = \sqrt {1 - \dfrac{4}{9}}$
$\Rightarrow \,\,\sin \theta = \sqrt {\dfrac{{9 - 4}}{9}}$
On simplification, we get
$\Rightarrow \,\,\sin \theta = \dfrac{{\sqrt 5 }}{3}$
Substitute the value of $\sin \alpha$ and $\cos \alpha$ in equation (2), then
$\Rightarrow \,\,\tan \alpha = \dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{\dfrac{2}{3}}}$
Or
$\Rightarrow \,\,\tan \alpha = \dfrac{{\sqrt 5 }}{3} \times \dfrac{3}{2}$
On simplification, we get
$\Rightarrow \,\,\tan \alpha = \dfrac{{\sqrt 5 }}{2}$
Where $\alpha = \arccos \left( {\dfrac{2}{3}} \right)$, then
$\Rightarrow \,\,\tan \left( {\arccos \left( {\dfrac{2}{3}} \right)} \right) = \dfrac{{\sqrt 5 }}{2}$

Hence, the value of $\tan \left( {\arccos \left( {\dfrac{2}{3}} \right)} \right)$ is $\dfrac{{\sqrt 5 }}{2}$.

Note: Here the given question belongs to the topic trigonometry. In the question we have the word tan which means it is tangent trigonometry ratio. Here we must know the definition of trigonometric ratios and standard identity. By the table of trigonometric ratios for the standard angles we simplify the given trigonometric function and hence we obtain the required result for the given question.