Question

How do you simplify $\dfrac{\sec x-\cos x}{\tan x}$ ?

Answer 1

We separate the terms in the numerator to make two separate fractions. We convert the secant trigonometric function $\sec x$ in the numerator of the first term into tangent and sine to cancel out the tangent. We convert the tangent trigonometric function $\tan x$ into sine and cosine using $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, We simplify and use Pythagorean trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ .

Complete step by step solution:
We know that we convert tangent trigonometric function into sin and cosine using the following formula
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
We are given in the following trigonometric expression to simplify
$\dfrac{\sec x-\cos x}{\tan x}$
Let us separate the two terms of the numerator into two separate fractions. We have;
$\Rightarrow \dfrac{\sec x}{\tan x}-\dfrac{\cos x}{\sin x}$
We know that cosine and secant trigonometric functions are reciprocal of each other which means $\sec x=\dfrac{1}{\cos x}$. We multiply $\sin x$ in the numerator and denominator to have $\sec x=\dfrac{\sin x}{\cos x\cdot \sin x}=\dfrac{\dfrac{\sin x}{\cos x}}{\cos x}=\dfrac{\tan x}{\cos x}$. We use this expression in the above step to have;
$\Rightarrow \dfrac{\dfrac{\tan x}{\sin x}}{\tan x}-\dfrac{\cos x}{\tan x}$
We cancel out $\tan x$ in the numerator and denominator in the above step to have;

& \Rightarrow \dfrac{\dfrac{\tan x}{\sin x}}{\tan x}-\dfrac{\cos x}{\tan x} \\\ & \Rightarrow \dfrac{1}{\sin x}-\dfrac{\cos x}{\tan x} \\\ \end{aligned}$$ We convert the tangent function in the second term to sine and cosine to have; $$\begin{aligned} & \Rightarrow \dfrac{1}{\sin x}-\dfrac{\cos x}{\dfrac{\sin x}{\cos x}} \\\ & \Rightarrow \dfrac{1}{\sin x}-\dfrac{{{\cos }^{2}}x}{\sin x} \\\ & \Rightarrow \dfrac{1-{{\cos }^{2}}x}{\sin x} \\\ \end{aligned}$$ We use Pythagorean trigonometric identity of sine and cosine ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ in the above step for $\theta =x$to have; $$\begin{aligned} & \Rightarrow \dfrac{{{\sin }^{2}}x}{\sin x} \\\ & \Rightarrow \dfrac{\sin x\cdot \sin x}{\sin x} \\\ & =\sin x \\\ \end{aligned}$$ The above expression is the required simplified expression .$$$$ **Note:** We can alternatively solve by directly converting the secant and tangent into sine and cosine as $$\begin{aligned} & \dfrac{\sec x-\cos x}{\tan x}\Rightarrow \dfrac{\dfrac{1}{\cos x}-\cos x}{\dfrac{\sin x}{\cos x}} \\\ & \Rightarrow \dfrac{\sec x-\cos x}{\tan x}=\dfrac{\dfrac{1-{{\cos }^{2}}x}{\cos x}}{\dfrac{\sin x}{\cos x}} \\\ \end{aligned}$$ We cancel out $\cos x$ in the above expression step to have $$\Rightarrow \dfrac{\sec x-\cos x}{\tan x}=\dfrac{1-{{\cos }^{2}}x}{\sin x}$$ We use Pythagorean trigonometric identity of sine and cosine ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ in the above step for $\theta =x$to have; $$\begin{aligned} & \Rightarrow \dfrac{\sec x-\cos x}{\tan x}=\dfrac{{{\sin }^{2}}x}{\sin x} \\\ & \Rightarrow \dfrac{\sec x-\cos x}{\tan x}=\sin x \\\ \end{aligned}$$ We note that here $x\ne \left( 2n+1 \right)\dfrac{\pi }{2}$ since $\tan x,\sec x$ are not defined for right angle or any odd multiples of right angle.