Relationships between the ratios
The six ratios are linked: cosec, sec, cot are reciprocals of sin, cos, tan, and tan A is sin A/cos A. One ratio gives all the others.
How the ratios are linked (ΔABC, ∠B = 90°)
| Relation | Formula | Why it holds |
|---|---|---|
| Reciprocal | cosec A = 1/sin A | sin A = BC/AC and cosec A = AC/BC |
| Reciprocal | sec A = 1/cos A | cos A = AB/AC and sec A = AC/AB |
| Reciprocal | cot A = 1/tan A | tan A = BC/AB and cot A = AB/BC |
| Quotient | tan A = sin A/cos A | (BC/AC) ÷ (AB/AC) = BC/AB |
| Quotient | cot A = cos A/sin A | (AB/AC) ÷ (BC/AC) = AB/BC |
Finding all ratios from one ratio (the k method)
- 1Write the given ratio as two sides with a common factor k. Example: sin A = 8/17 means BC = 8k and AC = 17k (k > 0).
- 2Draw a right triangle ABC with ∠B = 90° and write these lengths on it.
- 3★Find the third side by Pythagoras theorem.
- 4Write each required ratio from the figure. k cancels in every ratio.
- 5Check one answer with a relation, for example tan A = sin A/cos A.
★ key step
Given sin A, find the others
If sin A = 8/17, find cos A, tan A and sec A.
- sin A = opposite/hypotenuse = 8/17. Let BC = 8k and AC = 17k.
- AB² = AC² − BC² = 289k² − 64k² = 225k², so AB = 15k.
- cos A = AB/AC = 15k/17k = 15/17.
- tan A = BC/AB = 8k/15k = 8/15.
- sec A = 1/cos A = 17/15.
- Check: sin A/cos A = (8/17) ÷ (15/17) = 8/15 = tan A.
cos A = 15/17, tan A = 8/15, sec A = 17/15.
Evaluating an expression from tan θ
If 5 tan θ = 12, find the value of (sin θ + cos θ)/(sin θ − cos θ).
- tan θ = 12/5.
- Divide the top and the bottom by cos θ: (sin θ + cos θ)/(sin θ − cos θ) = (tan θ + 1)/(tan θ − 1).
- = (12/5 + 1)/(12/5 − 1)
- = (17/5)/(7/5) = 17/7.
17/7
Word problem: slope of a ghat road
A road up a hill rises 7 m for every 24 m it runs horizontally. θ is the angle between the road and the horizontal. Find sin θ and cos θ.
- tan θ = rise/run = 7/24. Let the rise be 7k and the run be 24k.
- Length along the road = √((7k)² + (24k)²) = √(49k² + 576k²) = √(625k²) = 25k.
- sin θ = 7k/25k = 7/25.
- cos θ = 24k/25k = 24/25.
sin θ = 7/25, cos θ = 24/25.
The steepness of a ghat road is often given as rise over run, which is tan θ. From it, engineers find the other ratios by the same k method you use here.
1 marks: “If sin A = 1/3, find cosec A.” cosec A = 1/sin A = 3.
2 marks: “If 7 cot A = 24, find sin A and sec A.” cot A = 24/7: adjacent = 24k, opposite = 7k, hypotenuse = 25k. sin A = 7/25, sec A = 25/24.
3 marks: “If sec θ = 5/4, find sin θ and tan θ, and the value of (1 − tan θ)/(1 + tan θ).” cos θ = 4/5: adjacent 4k, hypotenuse 5k, opposite 3k. sin θ = 3/5, tan θ = 3/4. (1 − 3/4)/(1 + 3/4) = (1/4)/(7/4) = 1/7.
Exactly one ratio in each pair has 'co': sin with cosec, cos with sec, tan with cot.