Tangent in Mathematics

Tangents are the most common concept of two vital topics of mathematics, geometry and trigonometry, having a different meaning and use in each one. Whether it's the sophisticated flair of a line to a curve in geometry, or how angles and ratios are brought to life in trigonometry, tangents are everywhere. So, let's explore how this simple yet profoundly insightful idea is central to everything, from circle theorems to the very structure of trigonometric identities.

1.0Tangents in Geometry

In geometry, the tangent is primarily interested in lines that are tangent to curves, such as circles, and in determining relationships between points, angles, and lengths. Tangents in geometry are used in various aspects of geometry with different meanings in different contexts, such as: 

Tangent to a Circle

A tangent to a circle is a line that is straight and touches the circle at a single point. The point where the tangent touches the circle is referred to as the point of tangency or the point of contact. Tangents are unique to a certain point, meaning no two tangents are drawn from a single point of contact. 

Theorems on Tangents for Circles

Tangents in geometry follow several theorems, especially when dealing with circles. These theorems form the basis of many proofs and constructions in other topics of geometry:  

The Tangent-Radius Theorem:

A tangent to a circle is always at right angles to the radius drawn to the point of contact.

If a line is touching a circle at point P and O is the centre, then OA is tangent.

Two Tangents from an External Point:

From any external point of a circle, two tangents can be drawn exactly, and both are of equal length.

If PA and PB are tangents from P to a circle with points of contact A and B, then PA = PB.

Tangent to a Curve

A tangent to a curve is a straight line that touches the curve at one and only one point and does not cut across it. The tangent line best approximates the curve linearly at the point. The most important property of the tangent is that it is the instantaneous rate of change (slope) of the curve at the point of tangency.

2.0Tangent Formula and Derivation of Tangent Formula

The tangent formula is derived from the tangent secant theorem, which states, if given a circle with centre O, PQ be a tangent to this circle with external point P, and PRS be the secant to the same circle with R and S being the points on the circle. Then, according to this theorem, the tangent formula can be written as: 

$P{Q}^2=PS\times PR$

Derivation of Tangent Formula

As per the figure given below, where O is the centre of the circle, PRS is a secant of the circle, and PQ is the tangent to the circle. A line OA is drawn perpendicular to RS. Now, Join OR, OP and OQ.

From the construction we know, OP ⟂ PQ, hence, 

RA = AS …… (1) (A perpendicular drawn from the centre of the circle on the chord bisects the chord)

PR = (PA – RA) and PS = (PA + AS) 

Now, Multiply PR with PS

PR × PS = (PA – RA) (PA + AS)

⇒ PR × PS = (PA – RA) (PA + RA) [From equation 1]

⇒ PR × PS = PA² – RA² ….(2)

In △ POA, by Pythagoras' theorem

OP² = OA² + PA²

⇒ PA² =  OP² – OA²

From equation 2 

⇒ PR × PS = OP² – OA² – RA²

⇒ PR × PS = OP² – (OA² + RA²) …..(3)

Now, similarly in △ ROA

OR² = OA² + RA²

⇒ RA² = OR² – OA² 

From equation 3 

⇒ PR × PS = OP² – (OA² + OR² – OA²)

⇒ PR × PS = OP² – OA² – OR² + OA²)

⇒ PR × PS = OP² – OR² …. (4) 

Since OR = OQ (Radii of the circle) 

Thus, equation 4 becomes

PR × PS = OP² – OQ² ….. (5)

In △ POQ

OP² = OQ² + PQ² 

⇒ PQ² = OP² – OQ²

From equation 5

PQ² = PR × PS 

3.0Trigonometric Interpretation of Tangents

Now that we know about the tangents in geometry, let’s take a quick look at the tangents of trigonometry. In trigonometry, the term tangent is related to the right-angled triangle. Tangents are the functions that play a critical role in finding the angular measurement of these triangles, and are known as the trigonometric ratios. The formula to find these tangents is: 

$tan\theta =\frac{\text{Perpendicular}}{\text{ Base}}$

Here is the angle between the perpendicular and the base of the right-angled triangle. The numerical values for this tan are different for different angles. For example: 

4.0Equation of Tangent

The equation of a tangent to a curve at a given point can be determined using the concept of slope. It is represented by the derivative of the function. The general idea behind this is that the equation of the straight line with slope m passing through point (x₀, y₀) is: 

$y-y_0=m(x-x_0)$

For a curve defined by y = f(x), the slope of the tangent at a specific point x₀​ is given by the derivative:

$m=f^{\prime }(x_0)$

Hence, the equation of the tangent line to the curve y = f(x) can be written as: 

$y-y_0=f^{\prime }(x_0)(x-x_0)$

In a trigonometric context, this slope “m” can be related to an angle θ, which is the angle of inclination of the tangent line made with the positive x-axis. The relationship is given by:

$tan\theta =f^{\prime }(x_0)$

Hence, we conclude that the slope of a tangent line can be written as:

Slope of the tangent = $tan\theta$

5.0Application of Tangents

Tangents have significant applications in different fields of study, which include:

• Geometry: Building shapes and establishing theorems.
• Engineering: Creating roads, bridges, and railroads.
• Physics: Determining angles, forces, and motion rates.
• Computer Graphics: Smoothing curves and edges.
• Architecture: Building designs involving curves and circular structures.

6.0Solved Examples in Tangents

Problem 1: From a point P outside a circle, two tangents PA and PB are drawn to the circle. If PA = 7 cm, what is the length of PB? 

Solution: Given that PA and PB are two tangents from an external point P.

We know that from any external point of a circle, two tangents can be drawn exactly, and both are of equal length. Hence, 

PA = PB = 7cm 

Problem 2: Find the equation of the tangent to the curve y = x² at the point where x = 2. 

Solution: given that y = x² 

Put x = 2, y = 2² = 4. 

Point (2, 4) 

Now, y’(x) = 2x 

y’(x) = 4 

Using the equation of the tangent, we have; 

$y-y_0=y^{\prime }(x_0)(x-x_0)$

y-4=4(x-2)

y-4=4x-8

4x-y=4

Problem 3: If a tangent to a curve at a point makes an angle of 30 degrees with the x-axis, what is the slope of the tangent?

Solution: Given that the angle of tangent to a curve is 30°, we know: 

Slope of the tangent= $tan\theta$

Hence, 

Slope of the tangent= $tan30=\frac{1}{\sqrt{3}}$