Tenth Grade Mathematics: Fundamental Identities: Pythagorean and Quotient Identities
An equation is called an algebraic identity if it is satisfied by all possible values of the variables involved. Similarly, a trigonometric identity is an equation involving trigonometric ratios of an angle that holds true for all acute angles for which the functions are defined. While specific ratios vary with the angle, the fundamental identities represent permanent, invariant geometrical relationships between the ratios themselves. Derived directly from the Pythagorean theorem, these master identities form the computational engine of analytical trigonometry.
The First Fundamental Identity: sin^2(theta) + cos^2(theta) = 1
Consider a right-angled triangle ABC with the right angle at B, reference angle Angle A = theta, opposite side BC = p, adjacent side AB = b, and hypotenuse AC = h:
The Fundamental Pythagorean Proof:
By the Pythagorean theorem:
BC^2 + AB^2 = AC^2
p^2 + b^2 = h^2
Divide the entire equation by h^2 (the square of the hypotenuse):
(p^2 / h^2) + (b^2 / h^2) = (h^2 / h^2)
(p / h)^2 + (b / h)^2 = 1
By definition of the primary ratios: p / h = sin(theta) and b / h = cos(theta). Substituting these into the equation yields the premier identity of trigonometry:
Theorem (First Pythagorean Identity):
sin^2(theta) + cos^2(theta) = 1
for all 0° <= theta <= 90°.
Transposed Variants:
1. sin^2(theta) = 1 - cos^2(theta) ===> sin(theta) = sqrt(1 - cos^2(theta))
2. cos^2(theta) = 1 - sin^2(theta) ===> cos(theta) = sqrt(1 - sin^2(theta))
The Second Fundamental Identity: 1 + tan^2(theta) = sec^2(theta)
Return to the fundamental Pythagorean relation:
p^2 + b^2 = h^2
Divide the entire equation by b^2 (the square of the adjacent side, for b != 0):
(p^2 / b^2) + (b^2 / b^2) = (h^2 / b^2)
(p / b)^2 + 1 = (h / b)^2
Since p / b = tan(theta) and h / b = sec(theta):
Theorem (Second Pythagorean Identity):
1 + tan^2(theta) = sec^2(theta)
for all 0° <= theta < 90°.
Transposed Variants:
1. sec^2(theta) - tan^2(theta) = 1
2. tan^2(theta) = sec^2(theta) - 1
3. (sec(theta) - tan(theta)) * (sec(theta) + tan(theta)) = 1
sec(theta) - tan(theta) = 1 / (sec(theta) + tan(theta))
The third variant is an exceptionally powerful shortcut: sec(theta) - tan(theta) and sec(theta) + tan(theta) are mutual reciprocals!
The Third Fundamental Identity: 1 + cot^2(theta) = cosec^2(theta)
Now divide the Pythagorean relation by p^2 (the square of the opposite side, for p != 0):
(p^2 / p^2) + (b^2 / p^2) = (h^2 / p^2)
1 + (b / p)^2 = (h / p)^2
Since b / p = cot(theta) and h / p = cosec(theta):
Theorem (Third Pythagorean Identity):
1 + cot^2(theta) = cosec^2(theta)
for all 0° < theta <= 90°.
Transposed Variants:
1. cosec^2(theta) - cot^2(theta) = 1
2. cot^2(theta) = cosec^2(theta) - 1
3. cosec(theta) - cot(theta) = 1 / (cosec(theta) + cot(theta))
Summary Chart of Fundamental Identities
+-----------------------+---------------------------------+-------------------------+
| Category | Primary Identity | Alternative Forms |
+-----------------------+---------------------------------+-------------------------+
| Pythagorean 1 | sin^2(x) + cos^2(x) = 1 | sin^2 = 1 - cos^2 |
| | | cos^2 = 1 - sin^2 |
+-----------------------+---------------------------------+-------------------------+
| Pythagorean 2 | 1 + tan^2(x) = sec^2(x) | sec^2 - tan^2 = 1 |
| | | tan^2 = sec^2 - 1 |
+-----------------------+---------------------------------+-------------------------+
| Pythagorean 3 | 1 + cot^2(x) = cosec^2(x) | cosec^2 - cot^2 = 1 |
| | | cot^2 = cosec^2 - 1 |
+-----------------------+---------------------------------+-------------------------+
| Quotient | tan(x) = sin(x) / cos(x) | cot(x) = cos(x)/sin(x) |
+-----------------------+---------------------------------+-------------------------+
| Reciprocal | sin(x) * cosec(x) = 1 | cos(x) * sec(x) = 1 |
| | tan(x) * cot(x) = 1 | |
+-----------------------+---------------------------------+-------------------------+
Chapter Exercises
Exercise 1: Express the trigonometric ratios sin(A), sec(A), and tan(A) purely in terms of cot(A).
Exercise 2: Evaluate the exact value of: 9sec^2(A) - 9tan^2(A), and evaluate (1 + tan(theta) + sec(theta))*(1 + cot(theta) - cosec(theta)).
Exercise 3: If sec(theta) + tan(theta) = p, find the values of sec(theta) and tan(theta) in terms of p, and hence deduce the value of sin(theta).
Solutions and Explanations
Solution 1: We express all ratios in terms of cot(A). From the third Pythagorean identity: 1 + cot^2(A) = cosec^2(A) ==> cosec(A) = sqrt(1 + cot^2(A)). Since sin(A) = 1 / cosec(A): sin(A) = 1 / sqrt(1 + cot^2(A)). For tan(A), use the reciprocal identity: tan(A) = 1 / cot(A). For sec(A), use sec(A) = sqrt(1 + tan^2(A)) = sqrt(1 + 1/cot^2(A)) = sqrt((cot^2(A) + 1) / cot^2(A)) = sqrt(1 + cot^2(A)) / cot(A).
Solution 2: For 9sec^2(A) - 9tan^2(A), factor out 9: 9(sec^2(A) - tan^2(A)). Since sec^2(A) - tan^2(A) = 1: 9(1) = 9. For the second expression: (1 + tan(theta) + sec(theta))(1 + cot(theta) - cosec(theta)). Convert everything to sines and cosines: (1 + sin/cos + 1/cos) * (1 + cos/sin - 1/sin) = [(cos + sin + 1)/cos] * [(sin + cos - 1)/sin] = [(sin + cos) + 1] * [(sin + cos) - 1] / (sin * cos). Apply difference of squares to numerator: [(sin + cos)^2 - 1^2] / (sin * cos) = [sin^2 + cos^2 + 2sincos - 1] / (sin * cos). Since sin^2 + cos^2 = 1: [1 + 2sincos - 1] / (sin * cos) = (2sin*cos) / (sin * cos) = 2.
Solution 3: We are given sec(theta) + tan(theta) = p (Equation 1). From the identity sec^2(theta) - tan^2(theta) = 1, we have (sec(theta) + tan(theta))(sec(theta) - tan(theta)) = 1. Substitute p: p * (sec(theta) - tan(theta)) = 1 ==> sec(theta) - tan(theta) = 1/p (Equation 2). Add Equation 1 and Equation 2: 2sec(theta) = p + 1/p = (p^2 + 1)/p ==> sec(theta) = (p^2 + 1) / (2p). Subtract Equation 2 from Equation 1: 2tan(theta) = p - 1/p = (p^2 - 1)/p ==> tan(theta) = (p^2 - 1) / (2p). Now find sin(theta): sin(theta) = tan(theta) / sec(theta) = [(p^2 - 1)/(2p)] / [(p^2 + 1)/(2*p)] = (p^2 - 1) / (p^2 + 1).