Divide Pythagoras by AB²

The idea

1 + tan² A = sec² A for 0° ≤ A < 90°, obtained by dividing AB² + BC² = AC² by AB²; it is not valid at 90°, where tan A and sec A are not defined.

Divide by 4² this time

In the 3-4-5 triangle, 3² + 4² = 5². Now divide every term by 4², the square of the side AB next to the angle A.

The side 4 is AB, the adjacent side. The side 3 is BC and the side 5 is AC.

What does the left side become?

• 1 + 9/16
• 1 + 3/4
• 9/25 + 16/25

Divide by AB²

Divide every term by AB², at two angles. Then push A all the way to 90° and see what happens to AB.

A second identity

1 + tan² A = sec² A for 0° ≤ A < 90°.

Divide AB² + BC² = AC² by AB²: (AB/AB)² + (BC/AB)² = (AC/AB)² i.e. 1 + tan² A = sec² A. It is true at A = 0° (1 + 0 = 1). At A = 90° the side AB is 0, so tan A and sec A are not defined. So the identity holds for 0° ≤ A < 90°.

Notes

1 + tan² A = sec² A for 0° ≤ A < 90°. It is AB² + BC² = AC² divided by AB². At 90° the divisor AB² is 0, so the identity does not apply.

Check yourself

tan A = 3/4. Use 1 + tan² A = sec² A to find sec² A, as a decimal to two places.

Answer: 1.56

sec² A = 1 + tan² A = 1 + 9/16 = 25/16, about 1.56. (So sec A = 5/4, as in the 3-4-5 triangle.)

sec A = 2. Find tan² A.

Answer: 3

tan² A = sec² A − 1 = 4 − 1 = 3. (This is A = 60°, where tan 60° = √3.)

Is 1 + tan² A = sec² A true at A = 0°?

Why does the identity 1 + tan² A = sec² A not include A = 90°?

• Yes: 1 + 0 = 1, and sec 0° = 1. — correct. Yes! tan 0° = 0 and sec 0° = 1, so both sides are 1.
• No, because tan 0° = 0.. tan 0° = 0 is allowed. Then the left side is 1 + 0 = 1, and sec² 0° = 1 too.
• It is not defined at 0°.. tan 0° and sec 0° are both defined (0 and 1). The trouble is at 90°.
• At 90° the side AB is 0, so tan 90° and sec 90° are not defined. — correct. Yes! We divided by AB², and AB² = 0 at 90°.
• Because 90° is not a whole number.. 90° is a whole number. The problem is dividing by 0.
• Because sec 90° = 1.. sec 90° has 0 at the bottom (1 / cos 90°), so it has no value.