Question

How do you find the exact value of $\sec \left( {\arctan \left( {\dfrac{{12}}{5}} \right)} \right)?$

Answer 1

In the above-given problem first we try to remove the inverse function and then applying the trigonometric formula in which the relationship between tangent and secant is given. When we get such a type of relation, we can put the value of tangent so that the required value of secant can be obtained. By keeping the above steps in mind, we can proceed to solve the problem.

(i). $\sec \varphi = \sqrt {1 + {{\tan }^2}\varphi }$

Complete step-by-step solution:
First, we are writing the above given expression as following,
$= \sec \left( {\arctan \left( {\dfrac{{12}}{5}} \right)} \right)$
Let $\arctan \left( {\dfrac{{12}}{5}} \right)$ be as any angle ‘$\theta$’ and it can be written as following,
$\Rightarrow \theta = \arctan \left( {\dfrac{{12}}{5}} \right)$
Now, it can also be written as following,
$\Rightarrow \theta = ta{n^{ - 1}}\left( {\dfrac{{12}}{5}} \right)$
Taking ${\tan ^{ - 1}}$ to the left-hand side, we get
$\Rightarrow \tan \theta = \left( {\dfrac{{12}}{5}} \right)$
As we know by above trigonometric formula in which the relationship between secant and tangent exist. Therefore, by applying the formula we can write it as,
$\Rightarrow \sec \theta = \sqrt {1 + {{\tan }^2}\theta }$
Putting value of $\tan \theta$ in the above equation, we get
$\Rightarrow \sec \theta = \sqrt {1 + {{\left( {\dfrac{{12}}{5}} \right)}^2}}$
Removing square of $\left( {\dfrac{{12}}{5}} \right)$ and it can be written as following,
$\Rightarrow \sec \theta = \sqrt {1 + \left( {\dfrac{{12}}{5} \times \dfrac{{12}}{5}} \right)}$
By multiplying under-root term, we get
$\Rightarrow \sec \theta = \sqrt {1 + \left( {\dfrac{{144}}{{25}}} \right)}$
Simplifying under-root term by taking L.C.M, we get
$\Rightarrow \sec \theta = \sqrt {\left( {\dfrac{{25 + 144}}{{25}}} \right)}$
Adding under-root term, we get
$\Rightarrow \sec \theta = \sqrt {\left( {\dfrac{{169}}{{25}}} \right)}$
Taking-out the under-root value, we get
$\Rightarrow \sec \theta = \dfrac{{13}}{5}$
Replacing ‘$\theta$’ with actual given function, we get
$\Rightarrow \sec \left( {\arctan \left( {\dfrac{{12}}{5}} \right)} \right) = \dfrac{{13}}{5}$
So, we have obtained the required value as asked in the problem.

Note: when we are solving a problem in which the arc of any trigonometric ratios is given, we should try to first write it in the inverse form so that we can easily apply any trigonometric formula as per requirement. There are many formulas in trigonometry but we have to choose such one in which the relation between two or more than two trigonometric ratios exists as the same as given in the expression of the problem.