Read tangent branches and keep the undefined inputs excluded
Use tangent as sine divided by cosine · Recognise poles at 90° and 270° within a full turn · Use period 180° without inventing a finite amplitude
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Full-turn sine and cosine
- Function domains and graph reading
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Use tangent as sine divided by cosine
- Recognise poles at 90° and 270° within a full turn
- Use period 180° without inventing a finite amplitude
A quotient with missing inputs
tanθ=sinθ/cosθ wherever cosine is non-zero. On 0°≤θ≤360°, tangent is undefined at 90° and 270°. It has zeros at 0°,180°,360° and fundamental period 180°. Its range is all real numbers, so it has no finite amplitude. The excluded inputs repeat every 180°.
Draw separate branches
As an input approaches 90° or 270°, values can become arbitrarily large in magnitude. The graph must not connect across either vertical asymptote. A curve leaving the displayed vertical window has not reached a domain endpoint; a finite plotting range cannot show its full unbounded behaviour.
Sources for this lesson
Pause and explain
Should a tangent curve join across θ=90°?
Worked example
State tangent zeros and excluded inputs on 0°≤θ≤360°.
Show the worked solution
- Zeros occur where sine is zero and cosine is non-zero.
- Cosine zero makes the quotient undefined.
Answer Zeros 0°,180°,360°; excluded inputs 90°,270°.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find tan0°, tan45°, tan135° and describe tan90°.
Give me a hint
Use sine divided by cosine and the quadrant.
Compare my reasoning
- The defined values are 0,1,−1.
- At 90° cosine is zero, so tangent is undefined.
0; 1; −1; undefined.
Look for these in your work
- Use quadrant signs
- Check the denominator
Solve tanθ=1 on 0°≤θ≤360°.
Give me a hint
Use the 180° period from the 45° value.
Compare my reasoning
- 45° is one solution.
- Add 180° to obtain 225°; the other repetitions lie outside the interval.
45° and 225°.
Look for these in your work
- Use the fundamental period
- Respect the interval
A student draws a straight connector between tangent values just below and just above 90°. Explain why this is not a valid tangent graph.
Give me a hint
Check the excluded input and the branch behaviour.
Compare my reasoning
- The function has no value at 90°.
- The two branches approach opposite unbounded values and cannot be joined across that pole.
The connector falsely includes an undefined input and hides the discontinuity.
Look for these in your work
- Exclude the pole
- Keep the branches separate
Common mistakes
- An asymptote is not a vertical piece of the function.
- Tangent has no finite amplitude.
What can you explain now?
Should a tangent curve join across θ=90°?
Compare with the explanation
No; tangent is undefined there.
Connecting the branches would falsely suggest a defined continuous value at the pole.
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Make it stick
Should a tangent curve join across θ=90°?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.