Answer :

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) + cos⁡²(x) = 1. Therefore, 1 - sin⁡²(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.