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Question: How do you find the exact value of \(\sec \left( \arctan \left( -\dfrac{3}{5} \right) \right)\)?...

How do you find the exact value of sec(arctan(35))\sec \left( \arctan \left( -\dfrac{3}{5} \right) \right)?

Explanation

Solution

We will look at the definition of the inverse trigonometric function. We will use the inverse trigonometric function to write the value of the angle. Then we will use the definition of the tangent function to obtain the opposite and adjacent side values. Using the Pythagoras theorem, we will find the value of the hypotenuse. Then we will use the definition of the secant function to obtain the value of the given function.

Complete step-by-step solution:
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 tangent function inside the bracket first.
Let x=arctan(35)x=\arctan \left( -\dfrac{3}{5} \right). Therefore, we have tanx=35\tan x=-\dfrac{3}{5}. The definition of the tangent function in a right angled triangle is given as tanx=OppositeAdjacent\tan x=\dfrac{\text{Opposite}}{\text{Adjacent}}.

Therefore, we get the adjacent side value as 5 and the opposite side value as 3-3.
Now, using the Pythagoras theorem in this right angled triangle, we get the following,
Hypotenuse=(Opposite)2+(Adjacent)2\text{Hypotenuse}=\sqrt{{{\left( \text{Opposite} \right)}^{2}}+{{\left( \text{Adjacent} \right)}^{2}}}
Substituting the values, we get
Hypotenuse=(3)2+(5)2 Hypotenuse=9+25 Hypotenuse=34 \begin{aligned} & \text{Hypotenuse}=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 5 \right)}^{2}}} \\\ & \Rightarrow \text{Hypotenuse}=\sqrt{9+25} \\\ & \therefore \text{Hypotenuse}=\sqrt{34} \\\ \end{aligned}
The secant function is defined as secx=1cosx=HypotenuseAdjacent\sec x=\dfrac{1}{\cos x}=\dfrac{\text{Hypotenuse}}{\text{Adjacent}}. Now, we have the following,
sec(arctan(35))=secx sec(arctan(35))=HypotenuseAdjacent \begin{aligned} & \sec \left( \arctan \left( -\dfrac{3}{5} \right) \right)=\sec x \\\ & \therefore \sec \left( \arctan \left( -\dfrac{3}{5} \right) \right)=\dfrac{\text{Hypotenuse}}{\text{Adjacent}} \\\ \end{aligned}
Substituting the value of the hypotenuse and the adjacent side, we get
sec(arctan(35))=345\sec \left( \arctan \left( -\dfrac{3}{5} \right) \right)=\dfrac{\sqrt{34}}{5}
Thus, we have obtained the value of the given function.

Note: We should be thoroughly familiar with the concept of trigonometric functions using the right angled triangle. It is very useful for such types of questions. We should understand the meaning of the inverse trigonometric function. Even though we did not find the exact value of the angle using the inverse trigonometric function, we used the concept and its definition in the solution.