Divide Pythagoras by BC²
Divide AB² + BC² = AC² by BC² and the identity cot² A + 1 = cosec² A appears.
Divide by 3² this time
In the 3-4-5 triangle, 3² + 4² = 5². Now divide every term by 3², the square of the side BC opposite to the angle A.
The side 4 is AB (adjacent), the side 3 is BC (opposite) and the side 5 is AC (hypotenuse).
What does the left side become?
What this lesson covers
The idea
cot² A + 1 = cosec² A for 0° < A ≤ 90°, obtained by dividing AB² + BC² = AC² by BC²; it is not valid at 0°, where cot A and cosec A are not defined.
Divide by 3² this time
In the 3-4-5 triangle, 3² + 4² = 5². Now divide every term by 3², the square of the side BC opposite to the angle A.
The side 4 is AB (adjacent), the side 3 is BC (opposite) and the side 5 is AC (hypotenuse).
What does the left side become?
- (4/3)² + 1
- 4/3 + 1
- (3/4)² + 1
Divide by BC²
Divide every term by BC², at two angles. Then pull A all the way down to 0° and see what happens to BC.
A third identity
cot² A + 1 = cosec² A for 0° < A ≤ 90°.
Divide AB² + BC² = AC² by BC²: (AB/BC)² + (BC/BC)² = (AC/BC)² i.e. cot² A + 1 = cosec² A. At A = 0° the side BC is 0, so cot A and cosec A are not defined. So the identity holds for 0° < A ≤ 90°.
Notes
cot² A + 1 = cosec² A for 0° < A ≤ 90°. It is AB² + BC² = AC² divided by BC². At 0° the divisor BC² is 0, so the identity does not apply.
Check yourself
cot A = 4/3. Use cot² A + 1 = cosec² A to find cosec² A, as a decimal to two places.
Answer: 2.78
cosec² A = cot² A + 1 = 16/9 + 1 = 25/9, about 2.78. (So cosec A = 5/3, as in the 3-4-5 triangle.)
cosec A = 5/4. Find cot² A, as a decimal to two places.
Answer: 0.56
cot² A = cosec² A − 1 = 25/16 − 1 = 9/16, about 0.56. (So cot A = 3/4.)
Is cot² A + 1 = cosec² A true at A = 90°?
Why does the identity cot² A + 1 = cosec² A not include A = 0°?
- Yes: 0 + 1 = 1, since cot 90° = 0 and cosec 90° = 1. — correct. Yes! cot 90° = 0 and cosec 90° = 1, so both sides are 1.
- No, because cot 90° = 0.. cot 90° = 0 is allowed. The left side is 0 + 1 = 1, and cosec² 90° = 1 too.
- It is not defined at 90°.. cot 90° and cosec 90° are both defined (0 and 1). The trouble is at 0°.
- At 0° the side BC is 0, so cot 0° and cosec 0° are not defined. — correct. Yes! We divided by BC², and BC² = 0 at 0°.
- Because 0° is not an angle.. An angle of 0° is fine. The problem is dividing by 0.
- Because cos 0° = 0.. cos 0° = 1. The problem is that BC = 0, so we cannot divide by BC².