Question

How do you simplify the expression $\cos \left( \arctan \left( \dfrac{x}{5} \right) \right)$?

Answer 1

The expression given in the above question, which is $\cos \left( \arctan \left( \dfrac{x}{5} \right) \right)$, consists of the cosine of the inverse tangent function. Therefore, in order to simplify it we need to use the trigonometric identity which relates the cosine function with the tangent function. Therefore, we can use the trigonometric identity given as ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$. On substituting ${{\sec }^{2}}\theta =\dfrac{1}{{{\cos }^{2}}\theta }$ into the trigonometric identity, we will obtain the relation $\dfrac{1}{{{\cos }^{2}}\theta }=1+{{\tan }^{2}}\theta$, on taking the reciprocal of which we will obtain the relation ${{\cos }^{2}}\theta =\dfrac{1}{1+{{\tan }^{2}}\theta }$. From here we will get two values of $\cos \theta$ which will be $\cos \theta =\dfrac{1}{\sqrt{1+{{\tan }^{2}}\theta }}$ and $\cos \theta =\dfrac{-1}{\sqrt{1+{{\tan }^{2}}\theta }}$. Finally, on substituting $\theta =\arctan \left( \dfrac{x}{5} \right)$ into both of the obtained values of $\cos \theta$, we will obtain two simplified expressions.

Complete step by step solution:
Let us consider the expression given in the above question as
$\Rightarrow E=\cos \left( \arctan \left( \dfrac{x}{5} \right) \right)$
Since the above expression contains the cosine of the tangent function, we use the trigonometric identity given by
$\Rightarrow {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$
Now, we know that the secant function is equal to the reciprocal of the cosine function. Therefore, we can substitute ${{\sec }^{2}}\theta =\dfrac{1}{{{\cos }^{2}}\theta }$ in the above trigonometric identity to get
$\Rightarrow \dfrac{1}{{{\cos }^{2}}\theta }=1+{{\tan }^{2}}\theta$
Taking reciprocals of both the sides, we get
$\Rightarrow {{\cos }^{2}}\theta =\dfrac{1}{1+{{\tan }^{2}}\theta }$
On solving the above equation, we get
$\Rightarrow \cos \theta =\pm \dfrac{1}{\sqrt{1+{{\tan }^{2}}\theta }}$
Now, substituting $\theta =\arctan \left( \dfrac{x}{5} \right)$ in the above equation, we get
$\begin{aligned} & \Rightarrow \cos \left( \arctan \left( \dfrac{x}{5} \right) \right)=\pm \dfrac{1}{\sqrt{1+{{\tan }^{2}}\left( \arctan \left( \dfrac{x}{5} \right) \right)}} \\\ & \Rightarrow \cos \left( \arctan \left( \dfrac{x}{5} \right) \right)=\pm \dfrac{1}{\sqrt{1+{{\left[ \tan \left( \arctan \left( \dfrac{x}{5} \right) \right) \right]}^{2}}}} \\\ \end{aligned}$
Now, we know that $\arctan$ is equal to the inverse of the tangent function. Therefore, we can substitute $\tan \left( \arctan \left( \dfrac{x}{5} \right) \right)=\dfrac{x}{5}$ in the above equation to get

& \Rightarrow \cos \left( \arctan \left( \dfrac{x}{5} \right) \right)=\pm \dfrac{1}{\sqrt{1+{{\left( \dfrac{x}{5} \right)}^{2}}}} \\\ & \Rightarrow \cos \left( \arctan \left( \dfrac{x}{5} \right) \right)=\pm \dfrac{1}{\sqrt{1+\dfrac{{{x}^{2}}}{25}}} \\\ & \Rightarrow \cos \left( \arctan \left( \dfrac{x}{5} \right) \right)=\pm \dfrac{5}{\sqrt{{{x}^{2}}+25}} \\\ \end{aligned}$$ **Hence, the given expression is simplified as $$\pm \dfrac{5}{\sqrt{{{x}^{2}}+25}}$$.** **Note:** For solving these types of questions, we must be familiar with all of the trigonometric identities. Also, the notation for the inverse tangent function is different in this question. The inverse tangent function is usually denoted as ${{\tan }^{-1}}$, while it is denoted as $\arctan $ in the given question. We must remember that they both are the notations for the inverse tangent functions.