Question

How do you write an algebraic expression that is equivalent to $\sec \left( {{\tan }^{-1}}\left( 3x \right) \right)$ ?

Answer 1

Problems on trigonometric functions can be easily done by using the identity relations of trigonometric functions and the properties of inverse trigonometric functions. To solve the given problem, we first convert the given expression into the square root of the square and use the identity relationship of $\sec \theta$ and $\tan \theta$ . Then we apply the property of inverse tangent function to get the algebraic expression of the given trigonometric equation.

Complete step by step answer:
The trigonometric expression that we are given is
$\sec \left( {{\tan }^{-1}}\left( 3x \right) \right)$
To simplify the above expression, we first rewrite it as the square root of a square as shown below
$\Rightarrow \sqrt{{{\sec }^{2}}\left( {{\tan }^{-1}}\left( 3x \right) \right)}$
Here, we take the value of $\theta$ as ${{\tan }^{-1}}\left( 3x \right)$and rewrite the main expression as shown below
$\Rightarrow \sqrt{{{\sec }^{2}}\theta }$
Now, we can use the identity relationship between the trigonometric functions $\sec \theta$ and $\tan \theta$ as
${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$
We rewrite the expression using the above identity as
$\Rightarrow \sqrt{1+{{\tan }^{2}}\theta }$
Now, we substitute the value of $\theta$ in the above expression as shown below
$\Rightarrow \sqrt{1+{{\tan }^{2}}\left( {{\tan }^{-1}}\left( 3x \right) \right)}$
To omit the inverse trigonometric function, we use the inverse property of inverse tangent function which is $\tan \left( {{\tan }^{-1}}\varphi \right)=\varphi$
Where $\varphi =3x$
Hence, the main expression can be written as
$\Rightarrow \sqrt{1+{{\left( \tan \left( {{\tan }^{-1}}\left( 3x \right) \right) \right)}^{2}}}$
$\Rightarrow \sqrt{1+{{\left( 3x \right)}^{2}}}$
Further simplifying the above equation, we get
$\Rightarrow \sqrt{1+9{{x}^{2}}}$

Therefore, the algebraic expression that is equivalent to $\sec \left( {{\tan }^{-1}}\left( 3x \right) \right)$ is $\sqrt{1+9{{x}^{2}}}$

Note: We must be very careful about using the correct identity and property of the trigonometric functions as using inappropriate relations will lead to inaccurate answers in spite of a simplified version of the expression. Also, while solving we must avoid step jumps as much as possible for this type of problems, as it will increase the chances of making silly mistakes that will lead to incorrect answers of the given problem.