How do you evaluate ${{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$ ?
Answer
617.7k+ views
Hint: The term ${{\sec }^{-1}}$ in our expression is nothing but the inverse of $\sec \theta$. So first we equate our expression with a variable and then convert ${{\sec }^{-1}}$ into $\sec x$ by sending it to the other side of the equation. Since $\sec \theta$ is $\dfrac{1}{\cos \theta }$ , solving the given trigonometric quantity using $\cos \theta$would be easier. Find the value using the trigonometric values and then that $\theta$ will be the solution.
Complete step by step solution:
The given trigonometric expression is, ${{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$
Let us first equate it to a variable.
Let us now consider $x={{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$
Now apply $\sec \theta$ on both sides of the expression.
Then the expression will now be,
$\Rightarrow \sec x=\sec \left( {{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right) \right)$
Whenever there are a function and its inverse then,$f\left( {{f}^{-1}}\left( x \right) \right)=x$
On substituting the values in the above postulate, we get,
Hence our expression will be,
$\Rightarrow \sec x=\left( \dfrac{2}{\sqrt{3}} \right)$
Since $\sec \theta$ is $\dfrac{1}{\cos \theta }$ converts out the expression in the same way.
$\Rightarrow \dfrac{1}{\cos x}=\left( \dfrac{2}{\sqrt{3}} \right)$
On inverting the entire equation, we get,
$\Rightarrow \cos x=\dfrac{\sqrt{3}}{2}$
Using the trigonometric table for the values or using the right triangle and the Pythagoras theorem,
We know that $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$
Therefore, the value of $x$ where the $\cos$ value is $\dfrac{\sqrt{3}}{2}$ is when $x$ is equal to ${{30}^{\circ }}$ .
Hence $x={{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)={{30}^{\circ }}$ or $x=\dfrac{\pi }{6}$
Note: The inverse functions in trigonometry are also known as arc functions or anti trigonometric functions. They are majorly known as arc functions because they are most used to find the length of the arc needed to get the given or specified value. We can convert a function into an inverse function and vice versa. Always check when the trigonometric functions are given in degrees or radians. There is a lot of difference between both.$1{}^\circ \times \dfrac{\pi }{180}=0.017Rad$. Express everything in $\sin \theta$ or $\cos \theta$ to easily evaluate.
Complete step by step solution:
The given trigonometric expression is, ${{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$
Let us first equate it to a variable.
Let us now consider $x={{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$
Now apply $\sec \theta$ on both sides of the expression.
Then the expression will now be,
$\Rightarrow \sec x=\sec \left( {{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right) \right)$
Whenever there are a function and its inverse then,$f\left( {{f}^{-1}}\left( x \right) \right)=x$
On substituting the values in the above postulate, we get,
Hence our expression will be,
$\Rightarrow \sec x=\left( \dfrac{2}{\sqrt{3}} \right)$
Since $\sec \theta$ is $\dfrac{1}{\cos \theta }$ converts out the expression in the same way.
$\Rightarrow \dfrac{1}{\cos x}=\left( \dfrac{2}{\sqrt{3}} \right)$
On inverting the entire equation, we get,
$\Rightarrow \cos x=\dfrac{\sqrt{3}}{2}$
Using the trigonometric table for the values or using the right triangle and the Pythagoras theorem,
We know that $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$
Therefore, the value of $x$ where the $\cos$ value is $\dfrac{\sqrt{3}}{2}$ is when $x$ is equal to ${{30}^{\circ }}$ .
Hence $x={{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)={{30}^{\circ }}$ or $x=\dfrac{\pi }{6}$
Note: The inverse functions in trigonometry are also known as arc functions or anti trigonometric functions. They are majorly known as arc functions because they are most used to find the length of the arc needed to get the given or specified value. We can convert a function into an inverse function and vice versa. Always check when the trigonometric functions are given in degrees or radians. There is a lot of difference between both.$1{}^\circ \times \dfrac{\pi }{180}=0.017Rad$. Express everything in $\sin \theta$ or $\cos \theta$ to easily evaluate.
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