Answer:
See below for proof.
Step-by-step explanation:
Given trigonometric identity:
[tex]\dfrac{\sec(x)}{\sin(x)}-\dfrac{\sin(x)}{\cos(x)}=\cot(x)[/tex]
To prove the given trigonometric identity, manipulate the left side until it simplifies into a form equivalent to the right side, using known trigonometric identities and algebraic manipulations.
Begin by rewriting sec(x) in terms of cos(x) using the secant reciprocal identity:
[tex]\dfrac{\frac{1}{\cos(x)}}{\sin(x)}-\dfrac{\sin(x)}{\cos(x)}[/tex]
Now, express both fractions with a common denominator:
[tex]\dfrac{\frac{1}{\cos(x)}\cdot \cos(x)}{\sin(x)\cdot \cos(x)}-\dfrac{\sin(x)\cdot \sin(x)}{\cos(x)\cdot \sin(x)}\\\\\\\\\dfrac{1}{\sin(x)\cos(x)}-\dfrac{\sin^2(x)}{\sin(x)\cos(x)}[/tex]
Combine fractions:
[tex]\dfrac{1-\sin^2(x)}{\sin(x)\cos(x)}[/tex]
Recall the Pythagorean identity sin²(x) + cos²(x) = 1. Therefore, 1 - sin²(x) = cos²(x). Substituting this into the expression gives:
[tex]\dfrac{\cos^2(x)}{\sin(x)\cos(x)}[/tex]
Cancel the common factor cos(x):
[tex]\dfrac{\cos(x)}{\sin(x)}[/tex]
Finally, apply the cotangent quotient identity, cot(x) = cos(x) / sin(x):
[tex]\cot(x)[/tex]
Therefore, the left side is equal to cot(x), proving the identity.
