Question

How do you solve and find the value of $\tan \left( {{{\cos }^{ - 1}}\left( {\dfrac{6}{7}} \right)} \right)$ ?

Answer 1

Hint : The given function is trigonometric with respect to inverse function, in which to find the value of $\tan \left( {{{\cos }^{ - 1}}\left( {\dfrac{6}{7}} \right)} \right)$ , we must know all the basic inverse relations i.e., formulas of cos and tan functions and to find the given value we can use the relation of $\tan a = \dfrac{{\sin a}}{{\cos a}}$ and apply the formulas.
Formula used:
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$\sin \theta = \sqrt {1 - {{\cos }^2}\theta }$

Complete step by step solution:
Let us write the given function as:
$\tan \left( {{{\cos }^{ - 1}}\left( {\dfrac{6}{7}} \right)} \right)$
To find the value of the given function,
Let, $a = {\cos ^{ - 1}}\left( {\dfrac{6}{7}} \right) \in {Q_1}$
$\Rightarrow$ $\cos a = \dfrac{6}{7}$ ……………. 1
and $\tan a > 0$ .
We know that,
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Hence let us apply this relation and find the value of the function:
$\tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{{\sqrt {1 - {{\cos }^2}a} }}{{\cos a}}$
Substitute the obtained value of $\cos a$ as $\dfrac{6}{7}$ , as in equation 1 i.e.,
$\Rightarrow$ $\tan a = \dfrac{{\sqrt {1 - {{\left( {\dfrac{6}{7}} \right)}^2}} }}{{\dfrac{6}{7}}}$
$\Rightarrow$ $\tan a = \dfrac{{\sqrt {1 - \dfrac{{36}}{{49}}} }}{{\dfrac{6}{7}}}$
After simplifying the terms, we get:
$\Rightarrow$ $\tan a = \dfrac{{\sqrt {\dfrac{{49 - 36}}{{49}}} }}{{\dfrac{6}{7}}}$
$\Rightarrow$ $\tan a = \dfrac{{\sqrt {\dfrac{{13}}{{49}}} }}{{\dfrac{6}{7}}}$
Hence, after simplification, we get
$\tan a = \dfrac{\sqrt 13}{{6 }} = 0.6$
Therefore, the value of $\tan \left( {{{\cos }^{ - 1}}\left( {\dfrac{6}{7}} \right)} \right)$ is 0.6
So, the correct answer is “0.6”.

Note : The key point to solve any trigonometric function is that we must know all the formulas with respect to the related questions asked as it seems easy to solve the question, we must note the chart of all related functions with respect to the equation, and here are some of the formulas to be noted while solving:
${\sin ^2}\theta + {\cos ^2}\theta = 1$ , ${\tan ^2}\theta + 1 = {\sec ^2}\theta$ ,
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ .