These eleven original questions move from rewriting a single expression to full proofs and excluded values. They cover every lesson in trigonometric identities. Work in order on paper, then open each answer after you have your own final line.
Each question has a fully worked answer. Use the mistake log and retest queue to record any slip and come back to it after a few days. The triangle and bearings reasoning board and the non-calculator working trainer help when you want a second way to check.
Questions
1. Simplify 7sin²x + 7cos²x.
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7(sin²x + cos²x) = 7 × 1 = 7.
2. Write 3cos²x + 4sin²x in terms of sin x only.
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Replace cos²x with 1 − sin²x: 3(1 − sin²x) + 4sin²x = 3 − 3sin²x + 4sin²x = 3 + sin²x.
Check at x = 30°: original 3 × 0.75 + 4 × 0.25 = 3.25. Answer 3 + 0.25 = 3.25.
3. Given cos θ = 5/13 and θ is acute, find sin θ and tan θ.
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sin²θ = 1 − 25/169 = 144/169. θ is acute, so sin θ is positive: sin θ = 12/13.
tan θ = (12/13) ÷ (5/13) = 12/5.
4. Simplify (sec²x − 1) cos²x.
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sec²x − 1 = tan²x, so the expression is tan²x cos²x = (sin²x / cos²x) × cos²x = sin²x.
Valid where cos x ≠ 0.
5. Given tan x = 2, find the value of (3 sin x − cos x) / (sin x + 2 cos x).
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Divide top and bottom by cos x: (3 tan x − 1) / (tan x + 2) = (6 − 1) / (2 + 2) = 5/4.
Check with a triangle: sin x = 2/√5, cos x = 1/√5. Top = 6/√5 − 1/√5 = 5/√5. Bottom = 2/√5 + 2/√5 = 4/√5. The ratio is 5/4.
6. Given sin θ = −3/5 and 180° < θ < 270°, find cos θ and tan θ.
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cos²θ = 1 − 9/25 = 16/25, so cos θ = ±4/5. In the third quadrant cosine is negative: cos θ = −4/5.
tan θ = (−3/5) ÷ (−4/5) = 3/4.
7. Prove that tan²x − sin²x ≡ tan²x sin²x.
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LHS = sin²x / cos²x − sin²x = sin²x (1/cos²x − 1) = sin²x (1 − cos²x) / cos²x = sin²x × sin²x / cos²x = sin²x × tan²x = tan²x sin²x.
Check at x = 60°: LHS = 3 − 0.75 = 2.25. RHS = 3 × 0.75 = 2.25.
8. Prove that cos x / (1 − sin x) ≡ (1 + sin x) / cos x, and state the values of x in 0° ≤ x ≤ 360° for which it is valid.
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Multiply top and bottom of the left side by (1 + sin x): cos x (1 + sin x) / (1 − sin²x) = cos x (1 + sin x) / cos²x = (1 + sin x) / cos x.
Left side is undefined when sin x = 1, so x = 90°. The right side is undefined when cos x = 0, so x = 90° and 270°. The identity is valid except at x = 90° and 270°.
9. Prove that 1 / (sin x cos x) − cot x ≡ tan x, and state the excluded values in 0° ≤ x ≤ 360°.
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LHS = 1 / (sin x cos x) − cos x / sin x = [1 − cos²x] / (sin x cos x) = sin²x / (sin x cos x) = sin x / cos x = tan x.
The left side needs sin x ≠ 0 and cos x ≠ 0. Excluded values: x = 0°, 90°, 180°, 270°, 360°.
Check at x = 45°: 1 / 0.5 − 1 = 1, and tan 45° = 1.
10. A student writes (1 + sin²x) / (1 + sin x) = 1 + sin x. Test with x = 30°, decide if it is valid and show a correct simplification of (1 − sin²x) / (1 − sin x).
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At x = 30°: left side = 1.25 / 1.5 ≈ 0.833, but 1 + sin x = 1.5. Not valid: 1 + sin²x is not (1 + sin x)(1 + sin x).
For the correct version, 1 − sin²x = (1 − sin x)(1 + sin x), so (1 − sin²x) / (1 − sin x) = 1 + sin x, valid for sin x ≠ 1. Check at 30°: 0.75 / 0.5 = 1.5.
11. Prove that (1 + sec x) / (sin x + tan x) ≡ cosec x, and state the excluded values in 0° ≤ x ≤ 360°.
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Write in sin and cos: numerator = 1 + 1/cos x = (cos x + 1) / cos x. Denominator = sin x + sin x / cos x = sin x (cos x + 1) / cos x.
The fraction becomes [(cos x + 1) / cos x] ÷ [sin x (cos x + 1) / cos x] = (cos x + 1) / [sin x (cos x + 1)] = 1 / sin x = cosec x.
The original needs cos x ≠ 0, so 90° and 270° are excluded. The denominator sin x + tan x is zero when sin x = 0 or cos x = −1, giving 0°, 180° and 360°. Excluded values: 0°, 90°, 180°, 270°, 360°.
Check at x = 60°: 3 / 2.598 ≈ 1.155, and cosec 60° ≈ 1.155.
If you got these wrong
| Kind of error | Questions | Go to |
|---|---|---|
| Wrong identity or dropped square | 1 to 4 | Use a fundamental identity to rewrite an expression |
| Proof went both ways or stalled | 7, 8, 9, 11 | Prove an identity without assuming its conclusion |
| Long method, surds, or wrong quadrant sign | 5, 6 | Simplify an expression before substitution |
| Missing or incomplete excluded values | 8, 9, 11 | State excluded values in a trig identity |
| Illegal cancelling | 10 | Diagnose an invalid cancellation |
Where to go from here
When most of these feel secure, return to trigonometric identities and then move on to equations. If a question type keeps catching you, describe it to a teacher in online one-to-one Additional Mathematics tuition.