How do you find the exact value of $\sec \left( \arctan \left( -\dfrac{3}{5} \right) \right)$?
Answer
623.1k+ views
Hint: 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 $x=\arctan \left( -\dfrac{3}{5} \right)$. Therefore, we have $\tan x=-\dfrac{3}{5}$. The definition of the tangent function in a right angled triangle is given as $\tan x=\dfrac{\text{Opposite}}{\text{Adjacent}}$.
Therefore, we get the adjacent side value as 5 and the opposite side value as $-3$.
Now, using the Pythagoras theorem in this right angled triangle, we get the following,
$\text{Hypotenuse}=\sqrt{{{\left( \text{Opposite} \right)}^{2}}+{{\left( \text{Adjacent} \right)}^{2}}}$
Substituting the values, we get
$\begin{align}
& \text{Hypotenuse}=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 5 \right)}^{2}}} \\
& \Rightarrow \text{Hypotenuse}=\sqrt{9+25} \\
& \therefore \text{Hypotenuse}=\sqrt{34} \\
\end{align}$
The secant function is defined as $\sec x=\dfrac{1}{\cos x}=\dfrac{\text{Hypotenuse}}{\text{Adjacent}}$. Now, we have the following,
$\begin{align}
& \sec \left( \arctan \left( -\dfrac{3}{5} \right) \right)=\sec x \\
& \therefore \sec \left( \arctan \left( -\dfrac{3}{5} \right) \right)=\dfrac{\text{Hypotenuse}}{\text{Adjacent}} \\
\end{align}$
Substituting the value of the hypotenuse and the adjacent side, we get
$\sec \left( \arctan \left( -\dfrac{3}{5} \right) \right)=\dfrac{\sqrt{34}}{5}$
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.
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 $x=\arctan \left( -\dfrac{3}{5} \right)$. Therefore, we have $\tan x=-\dfrac{3}{5}$. The definition of the tangent function in a right angled triangle is given as $\tan x=\dfrac{\text{Opposite}}{\text{Adjacent}}$.
Therefore, we get the adjacent side value as 5 and the opposite side value as $-3$.
Now, using the Pythagoras theorem in this right angled triangle, we get the following,
$\text{Hypotenuse}=\sqrt{{{\left( \text{Opposite} \right)}^{2}}+{{\left( \text{Adjacent} \right)}^{2}}}$
Substituting the values, we get
$\begin{align}
& \text{Hypotenuse}=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 5 \right)}^{2}}} \\
& \Rightarrow \text{Hypotenuse}=\sqrt{9+25} \\
& \therefore \text{Hypotenuse}=\sqrt{34} \\
\end{align}$
The secant function is defined as $\sec x=\dfrac{1}{\cos x}=\dfrac{\text{Hypotenuse}}{\text{Adjacent}}$. Now, we have the following,
$\begin{align}
& \sec \left( \arctan \left( -\dfrac{3}{5} \right) \right)=\sec x \\
& \therefore \sec \left( \arctan \left( -\dfrac{3}{5} \right) \right)=\dfrac{\text{Hypotenuse}}{\text{Adjacent}} \\
\end{align}$
Substituting the value of the hypotenuse and the adjacent side, we get
$\sec \left( \arctan \left( -\dfrac{3}{5} \right) \right)=\dfrac{\sqrt{34}}{5}$
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.
Recently Updated Pages
Which of the following graphs shows the variation of class 12 physics CBSE

Draw a labelled diagram of the human male reproductive class 12 biology CBSE

Describe the experiment to compare the emf of two cells class 12 physics CBSE

What is standard hydrogen electrode

What is conventional current and electric current class 12 physics CBSE

2Bromopentane is treated with an alcoholic KOH solution class 12 chemistry CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

A member of Simon commission later became Prime Minister class 12 social science CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Give one example of a reptile that is viviparous class 12 biology CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

