Mathematics — AI-generated practice question

In △ABC, right-angled at B:

By Pythagoras theorem: $AB^2 + BC^2 = AC^2$ … (1)

Dividing (1) by $AC^2$:

$$\frac{AB^2}{AC^2} + \frac{BC^2}{AC^2} = 1$$

$$\left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = 1$$

Since $\cos A = \dfrac{AB}{AC}$ and $\sin A = \dfrac{BC}{AC}$:

$$\boxed{\sin^2 A + \cos^2 A = 1} \quad \text{...(2), valid for } 0° \leq A \leq 90°$$

---

Dividing (1) by $AB^2$:

$$1 + \frac{BC^2}{AB^2} = \frac{AC^2}{AB^2}$$

Since $\dfrac{BC}{AB} = \tan A$ and $\dfrac{AC}{AB} = \sec A$:

$$\boxed{1 + \tan^2 A = \sec^2 A} \quad \text{...(3), valid for } 0° \leq A < 90°$$

---

Dividing (1) by $BC^2$:

$$\frac{AB^2}{BC^2} + 1 = \frac{AC^2}{BC^2}$$

Since $\dfrac{AB}{BC} = \cot A$ and $\dfrac{AC}{BC} = \text{cosec } A$:

$$\boxed{\cot^2 A + 1 = \text{cosec}^2 A} \quad \text{...(4), valid for } 0° < A \leq 90°$$

Source: Chapter 8, Section 8.4 — Trigonometric Identities

---

• The examiner awards marks for: clearly stating Pythagoras theorem, correctly identifying the trig ratios after division (sin = BC/AC, cos = AB/AC, etc.), writing each identity in a box or clearly labelled, and noting the domain restriction for each.
• Identity (3) is undefined at A = 90° (tan and sec not defined there); identity (4) is undefined at A = 0° (cosec and cot not defined). Mentioning these restrictions earns the precision marks.
• Write $\sin^2 A + \cos^2 A = 1$ (not $\cos^2 A + \sin^2 A = 1$) as the question specifically asks for that form — both are acceptable but match the question.