MathsCBSE Class 10 · Introduction to Trigonometry

Relationships Between Trigonometric Ratios for Class 10

Q: Why can every trigonometric ratio be written using only sin and cos?

Cosec and sec are reciprocals of sin and cos, and tan and cot are quotients of sin and cos. So all six ratios are built from these two. That is why converting to sin and cos is the standard first step in simplification.

Q: How do I prove that tan A equals sin A divided by cos A?

In a right triangle, tan A is the opposite side over the adjacent side. Divide both by the hypotenuse. The top becomes sin A and the bottom becomes cos A, so tan A equals sin A over cos A.

Q: Is tan A always greater than sin A?

For an acute angle, yes. Tan A is sin A divided by cos A, and cos A lies between 0 and 1, so dividing by it makes the value larger.

Q: Are complementary angle relations like sin of 90 minus A part of this topic?

No. Trigonometric ratios of complementary angles are not in the 2026-27 syllabus.

Updated 14 October 2026 · Fact-checked

The six trigonometric ratios are linked. Cosec, sec and cot are the reciprocals of sin, cos and tan, and tan A equals sin A divided by cos A. The board asks you to use these links to find unknown ratios, write every ratio in terms of one ratio, simplify expressions and prove simple identities.

Each ratio pairs with its reciprocal to give 1, and tan and cot are built from sin and cos, so sin and cos are the two building blocks.

Illustration drawn for QuestionBuddy.

Understand Relationships Between Trigonometric Ratios

You have six trigonometric ratios, but they are not six separate pieces of information. Once you know one ratio of an acute angle, the others are fixed. The relationships in this topic tell you exactly how they are tied together.

Reciprocal relations. In $△ A B C$ , right-angled at $B$ , $sin A = \frac{B C}{A C}$ and $cosec A = \frac{A C}{B C}$ . One is the other turned upside down, so $cosec A = \frac{1}{s i n A}$ . In the same way $sec A = \frac{1}{c o s A}$ and $cot A = \frac{1}{t a n A}$ . So their products are 1: $sin A \cdot cosec A = 1$ , and so on.

Quotient relations. Start from $tan A = \frac{B C}{A B}$ and divide the top and bottom by $A C$ : $tan A = \frac{B C / A C}{A B / A C} = \frac{s i n A}{c o s A}$ . Turning this over gives $cot A = \frac{c o s A}{s i n A}$ . This is why sin and cos are the two building blocks: every other ratio can be written using only them.

Conditions. A fraction needs a non-zero denominator. So $cosec A$ and $cot A$ need $sin A \neq = 0$ (they fail at $0^{\circ}$ ), and $sec A$ and $tan A$ need $cos A \neq = 0$ (they fail at $9 0^{\circ}$ ). For an acute angle all six ratios exist and are positive.

How the ratios change with the angle. As $A$ grows from $0^{\circ}$ to $9 0^{\circ}$ , $sin A$ rises and $cos A$ falls, so $tan A = \frac{s i n A}{c o s A}$ rises. Reciprocals move the opposite way: $cosec A$ falls, $sec A$ rises and $cot A$ falls. Also, since $0 < cos A < 1$ for an acute angle, $tan A = \frac{s i n A}{c o s A}$ is always bigger than $sin A$ .

Connection to the rest of the chapter. Combined with $sin^{2} A + cos^{2} A = 1$ , these relations let you write every ratio in terms of any one ratio, for example $cos A = 1 - sin^{2} A$ and $tan A = \frac{s i n A}{1 - s i n ^{2} A}$ . Dividing $sin^{2} A + cos^{2} A = 1$ by $cos^{2} A$ gives $tan^{2} A + 1 = sec^{2} A$ , and dividing it by $sin^{2} A$ gives $1 + cot^{2} A = cosec^{2} A$ . These two forms let you start from tan or cot, for example $sec A = 1 + tan^{2} A$ and $cos A = \frac{1}{1 + t a n ^{2} A}$ . The reciprocal and quotient relations are also the first step in almost every identity proof: convert to sin and cos, take the LCM, simplify.

Relationships Between Trigonometric Ratios: formulas to remember

Reciprocal relations: $cosec A = \frac{1}{s i n A}, sec A = \frac{1}{c o s A}, cot A = \frac{1}{t a n A}$; Need sin A ≠ 0, cos A ≠ 0 and tan A ≠ 0 respectively.
Product forms: $sin A cosec A = 1, cos A sec A = 1, tan A cot A = 1$; Useful for quick simplification in MCQs.
Quotient relations: $tan A = \frac{s i n A}{c o s A}, cot A = \frac{c o s A}{s i n A}$; Proved by dividing numerator and denominator of the side ratio by the hypotenuse.
Mixed quotient forms: $\frac{s e c A}{cosec A} = tan A, sin A cot A = cos A, cos A tan A = sin A$; Follow directly from the reciprocal and quotient relations.
All ratios in terms of sin A (A acute): $cos A = 1 - sin^{2} A, tan A = \frac{s i n A}{1 - s i n ^{2} A}, cot A = \frac{1 - s i n ^{2} A}{s i n A}, sec A = \frac{1}{1 - s i n ^{2} A}, cosec A = \frac{1}{s i n A}$; Uses sin²A + cos²A = 1; take the positive root because A is acute.
Squared forms from sin²A + cos²A = 1: $1 + tan^{2} A = sec^{2} A, 1 + cot^{2} A = cosec^{2} A$; Divide sin²A + cos²A = 1 by cos²A for the first and by sin²A for the second.
All ratios in terms of tan A (A acute): $sin A = \frac{t a n A}{1 + t a n ^{2} A}, cos A = \frac{1}{1 + t a n ^{2} A}, sec A = 1 + tan^{2} A, cot A = \frac{1}{t a n A}, cosec A = \frac{1 + t a n ^{2} A}{t a n A}$; Uses sec²A = 1 + tan²A; take the positive root because A is acute.
Behaviour as A increases from 0° to 90°: $sin A ↑, tan A ↑, sec A ↑; cos A ↓, cot A ↓, cosec A ↓$; A reciprocal always moves opposite to its ratio.

How to solve Relationships Between Trigonometric Ratios questions

Almost every question on this topic is solved by turning everything into sin and cos, simplifying, and then turning back if needed.

1Note the angle condition (acute, or non-zero denominators) so you know every ratio used is defined and positive.
2Rewrite cosec, sec and cot as reciprocals, and tan and cot as quotients of sin and cos.
3If one ratio is given as a number, find sin and cos first (use $sin^{2} A + cos^{2} A = 1$ if needed, taking positive roots for acute angles).
4For an expression with the same power in numerator and denominator, divide top and bottom by a suitable ratio (such as $cos A$ or $cosec A$ ) to bring it to a single ratio.
5Combine fractions with an LCM, then use $sin^{2} A + cos^{2} A = 1$ to simplify.
6In a proof, work on one side only until it equals the other side.
7Check with a standard angle such as $4 5^{\circ}$ or $6 0^{\circ}$ before moving on.

Relationships Between Trigonometric Ratios in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Relationships between the ratios

Suppose your uncle is fixing a solar panel on the flat roof of your house. The panel is a straight board 2.5 m long. Its lower edge rests on the roof and its upper edge is held up by a vertical rod 1.5 m tall. The installer's chart lists cot A and cosec A, where A is the angle the panel makes with the roof.

Find the horizontal distance from the lower edge to the foot of the rod, then find cot A and cosec A using the reciprocal and quotient relations. The rod is then replaced by a 2.4 m rod, which is still shorter than the 2.5 m panel (panel length unchanged). Find the new horizontal distance, cot A and cosec A, and say whether cot A has increased or decreased.

Try it first, then show the solutionSolution

Right triangle: hypotenuse 2.5 m (panel), opposite side 1.5 m (rod).
Base $= 2. 5^{2} - 1. 5^{2} = 6.25 - 2.25 = 4 = 2$ m.
$sin A = \frac{1.5}{2.5} = \frac{3}{5}$ , $cos A = \frac{2}{2.5} = \frac{4}{5}$ .
$cot A = \frac{c o s A}{s i n A} = \frac{4/5}{3/5} = \frac{4}{3}$ .
$cosec A = \frac{1}{s i n A} = \frac{5}{3}$ .
New rod: base $= 2. 5^{2} - 2. 4^{2} = 6.25 - 5.76 = 0.49 = 0.7$ m.
New $cot A = \frac{0.7}{2.4} = \frac{7}{24}$ and new $cosec A = \frac{2.5}{2.4} = \frac{25}{24}$ .
Compare: $\frac{7}{24} < \frac{4}{3}$ , so cot A has decreased. This fits the rule: a taller rod makes angle A larger, so $cos A$ falls and $sin A$ rises, and $cot A = \frac{c o s A}{s i n A}$ falls.

Answer: Distance 2 m, $cot A = \frac{4}{3}$ , $cosec A = \frac{5}{3}$ . With the 2.4 m rod: distance 0.7 m, $cot A = \frac{7}{24}$ , $cosec A = \frac{25}{24}$ , so cot A decreases.

Memory hook: Steeper panel, smaller cot: cos falls while sin rises.

Tricks to remember Relationships Between Trigonometric Ratios

Short memory aids. Say each one aloud once, then use it on a question the same day.

One 'co' per pair
sin ↔ cosec, cos ↔ sec, tan ↔ cot. Each reciprocal pair has exactly one 'co' in it.
Use it when: Stops you pairing sin with sec by mistake.
Everything is sin and cos
tan = s/c, cot = c/s, sec = 1/c, cosec = 1/s.
Use it when: First line of any simplification or identity proof.
Reciprocals run the other way
If a ratio goes up as the angle grows, its reciprocal comes down.
Use it when: Ordering or comparison questions on increasing and decreasing ratios.

Relationships Between Trigonometric Ratios: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Using a product relation · 1 mark · MCQBoard level

If $sec A = \frac{5}{4}$ , where $A$ is acute, then the value of $sin A \cdot cot A$ is
(a) $\frac{3}{5}$
(b) $\frac{4}{5}$
(c) $\frac{5}{3}$
(d) $\frac{3}{4}$

Try it first, then show the solutionSolution

$sin A \cdot cot A = sin A \cdot \frac{c o s A}{s i n A} = cos A$ .
$cos A = \frac{1}{s e c A} = \frac{4}{5}$ .

Answer: (b) $\frac{4}{5}$

Example 2: Express every ratio in terms of sin A · 3 marks · Short answerBoard level

Express $cos A$ , $tan A$ , $sec A$ , $cosec A$ and $cot A$ in terms of $sin A$ , where $A$ is acute.

Try it first, then show the solutionSolution

From $sin^{2} A + cos^{2} A = 1$ , $cos A = 1 - sin^{2} A$ (positive since $A$ is acute). (1 mark)
$tan A = \frac{s i n A}{c o s A} = \frac{s i n A}{1 - s i n ^{2} A}$ and $sec A = \frac{1}{1 - s i n ^{2} A}$ . (1 mark)
$cosec A = \frac{1}{s i n A}$ and $cot A = \frac{1 - s i n ^{2} A}{s i n A}$ . (1 mark)

Answer: $cos A = 1 - sin^{2} A$ , $tan A = \frac{s i n A}{1 - s i n ^{2} A}$ , $sec A = \frac{1}{1 - s i n ^{2} A}$ , $cosec A = \frac{1}{s i n A}$ , $cot A = \frac{1 - s i n ^{2} A}{s i n A}$ .

Example 3: Using 1 + tan²A = sec²A · 3 marks · Short answerHarder board level

If $tan A = \frac{5}{12}$ , where $A$ is acute, find $sin A$ , $cos A$ and the value of $\frac{s e c A + cosec A}{s e c A - cosec A}$ .

Try it first, then show the solutionSolution

$sec^{2} A = 1 + tan^{2} A = 1 + \frac{25}{144} = \frac{169}{144}$ , so $sec A = \frac{13}{12}$ (positive, since $A$ is acute). (½ mark)
$cos A = \frac{1}{s e c A} = \frac{12}{13}$ . (½ mark)
$sin A = tan A cos A = \frac{5}{12} \times \frac{12}{13} = \frac{5}{13}$ , so $cosec A = \frac{13}{5}$ . (1 mark)
Expression $= \frac{\frac{13}{12} + \frac{13}{5}}{\frac{13}{12} - \frac{13}{5}} = \frac{\frac{65 + 156}{60}}{\frac{65 - 156}{60}} = \frac{221}{- 91} = - \frac{17}{7}$ . (1 mark)

Answer: $sin A = \frac{5}{13}$ , $cos A = \frac{12}{13}$ ; the expression equals $- \frac{17}{7}$ . It is negative because $cosec A > sec A$ when $tan A < 1$ .

Example 4: Proof using quotient and reciprocal relations · 3 marks · Short answerHarder board level

Prove that $tan A + cot A = sec A cosec A$ , where $A$ is acute. Verify for $A = 6 0^{\circ}$ .

Try it first, then show the solutionSolution

LHS $= \frac{s i n A}{c o s A} + \frac{c o s A}{s i n A}$ (quotient relations). (½ mark)
$= \frac{s i n ^{2} A + c o s ^{2} A}{s i n A c o s A}$ (LCM). (½ mark)
$= \frac{1}{s i n A c o s A}$ , since $sin^{2} A + cos^{2} A = 1$ . (½ mark)
$= \frac{1}{c o s A} \cdot \frac{1}{s i n A} = sec A cosec A$ = RHS. (½ mark)
Check at $6 0^{\circ}$ : LHS $= 3 + \frac{1}{3} = \frac{4}{3}$ . (½ mark)
RHS $= 2 \times \frac{2}{3} = \frac{4}{3}$ , so the result holds. (½ mark)

Answer: Proved; both sides equal $\frac{4}{3}$ at $6 0^{\circ}$ .

Example 5: All ratios from sin A, then an expression · 5 marks · Long answerHarder board level

If $sin A = \frac{2}{3}$ , where $A$ is acute:
(i) Find $cos A$ , $tan A$ , $cot A$ , $sec A$ and $cosec A$ .
(ii) Evaluate $\frac{t a n A + c o t A}{s e c A \cdot cosec A}$ .
(iii) Explain, using the relationships between the ratios, why the answer to (ii) is the same for every acute angle.

Try it first, then show the solutionSolution

$cos A = 1 - \frac{4}{9} = \frac{5}{3}$ . (½ mark)
$tan A = \frac{2/3}{5 /3} = \frac{2}{5}$ , $cot A = \frac{5}{2}$ . (1 mark)
$sec A = \frac{3}{5}$ , $cosec A = \frac{3}{2}$ . (½ mark)
$tan A + cot A = \frac{2}{5} + \frac{5}{2} = \frac{4 + 5}{2 5} = \frac{9}{2 5}$ . (1 mark)
$sec A cosec A = \frac{3}{5} \cdot \frac{3}{2} = \frac{9}{2 5}$ , so the value is $1$ . (½ mark)
In general $tan A + cot A = \frac{s i n ^{2} A + c o s ^{2} A}{s i n A c o s A} = \frac{1}{s i n A c o s A}$ . (1 mark)
And $sec A cosec A = \frac{1}{s i n A c o s A}$ , so the ratio is always 1. (½ mark)

Answer: (i) $cos A = \frac{5}{3}$ , $tan A = \frac{2}{5}$ , $cot A = \frac{5}{2}$ , $sec A = \frac{3}{5}$ , $cosec A = \frac{3}{2}$ . (ii) 1. (iii) Numerator and denominator both equal $\frac{1}{s i n A c o s A}$ .

Example 6: Reasoning with increasing and decreasing ratios · 3 marks · Short answerBoard level

$A$ and $B$ are acute angles with $A < B$ . State whether each statement is true or false, giving a reason.
(i) $cos A > cos B$
(ii) $sec A > sec B$
(iii) $tan A \cdot cot B < 1$

Try it first, then show the solutionSolution

(i) cos decreases as the angle increases, and $A < B$ , so $cos A > cos B$ : true. (1 mark)
(ii) $sec = \frac{1}{c o s}$ ; the larger cosine gives the smaller secant, so $sec A < sec B$ : false. (1 mark)
(iii) $tan A cot B = \frac{t a n A}{t a n B}$ ; tan increases, so $tan A < tan B$ and the ratio is less than 1: true. (1 mark)

Answer: (i) True (ii) False (iii) True

Where students lose marks in Relationships Between Trigonometric Ratios

The slips examiners see most often, and what to write instead.

Mistake: Pairing sin with sec
Students match by first letter or position instead of the 'one co' rule.
Wrong working
$sin A = \frac{3}{5}$ , so $sec A = \frac{5}{3}$ .
Right working
$cosec A = \frac{5}{3}$ ; $cos A = \frac{4}{5}$ , so $sec A = \frac{5}{4}$ .
Do this: sin–cosec, cos–sec, tan–cot: one 'co' per pair.
Usually costs: The whole answer in an MCQ; follow-on marks in longer questions.
Mistake: Multiplying instead of dividing when finding cot A from sin and cos
With two fractions on top of each other, students multiply them instead of dividing.
Wrong working
$cot A = \frac{c o s A}{s i n A}$ with $cos A = \frac{5}{3}$ , $sin A = \frac{2}{3}$ gives $cot A = \frac{5}{3} \cdot \frac{2}{3} = \frac{2 5}{9}$ .
Right working
$cot A = \frac{5}{3} \div \frac{2}{3} = \frac{5}{3} \times \frac{3}{2} = \frac{5}{2}$ .
Do this: Dividing by a fraction means multiplying by its reciprocal; write the division step.
Usually costs: ½ to 1 mark plus later steps.
Mistake: Thinking reciprocals change in the same direction
Students assume all ratios increase with the angle.
Wrong working
sin increases from 30° to 60°, so cosec 60° > cosec 30°.
Right working
$cosec 3 0^{\circ} = 2$ , $cosec 6 0^{\circ} = \frac{2}{3} < 2$ ; cosec decreases.
Do this: Bigger denominator, smaller fraction: a reciprocal moves opposite to its ratio.
Usually costs: 1 mark in MCQ or assertion-reason items.
Mistake: Writing a negative root for an acute angle
Taking $\pm$ out of habit from quadratic equations.
Wrong working
$cos A = \pm 1 - sin^{2} A$ .
Right working
For acute $A$ , $cos A = 1 - sin^{2} A$ (positive).
Do this: State 'A is acute, so all ratios are positive'.
Usually costs: ½ mark for an incomplete or incorrect final form.

Exam tips

When a proof mixes several ratios, your first line should convert them all to sin and cos; examiners give credit for that step.
For expressions like $\frac{s e c A + cosec A}{s e c A - cosec A}$ , dividing by one ratio to reach tan saves time and avoids square roots.
Write the condition (A acute, or sin A ≠ 0) whenever you divide by a ratio.
In assertion-reason items on which ratio is bigger, test the claim at $3 0^{\circ}$ and $6 0^{\circ}$ before choosing.
Verify any proof at $4 5^{\circ}$ or $6 0^{\circ}$ in rough work; it catches sign and reciprocal slips.

Practice questions on Relationships Between Trigonometric Ratios

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
If $sec A = \frac{17}{8}$ , where $A$ is acute, find $tan A$ and the value of $sin A \cdot cot A$ .
Hint
Find cos A from sec A; notice that sin A·cot A simplifies to one ratio.
Answer
$cos A = \frac{8}{17}$ , $sin A = \frac{15}{17}$ , so $tan A = \frac{15}{8}$ . $sin A cot A = cos A = \frac{8}{17}$ .
Q2 · 3 marks · Short answerHarder board level
Express $sec A$ and $cosec A$ in terms of $cot A$ , where $A$ is acute. Hence find $sec A$ and $cosec A$ when $cot A = \frac{7}{24}$ , and check that $sec^{2} A - tan^{2} A = 1$ .
Hint
Start from $cosec^{2} A = 1 + cot^{2} A$ , then use $sec A = \frac{cosec A}{c o t A}$ .
Answer
$cosec A = 1 + cot^{2} A$ and $sec A = \frac{cosec A}{c o t A} = \frac{1 + c o t ^{2} A}{c o t A}$ . With $cot A = \frac{7}{24}$ : $cosec A = 1 + \frac{49}{576} = \frac{25}{24}$ and $sec A = \frac{25}{24} \div \frac{7}{24} = \frac{25}{7}$ . Check: $tan A = \frac{24}{7}$ , so $sec^{2} A - tan^{2} A = \frac{625 - 576}{49} = 1$ .
Q3 · 3 marks · Short answerBoard level
Prove that $(sin A + cos A) (tan A + cot A) = sec A + cosec A$ , where $A$ is acute.
Hint
First show $tan A + cot A = \frac{1}{s i n A c o s A}$ .
Answer
$tan A + cot A = \frac{s i n ^{2} A + c o s ^{2} A}{s i n A c o s A} = \frac{1}{s i n A c o s A}$ . LHS $= \frac{s i n A + c o s A}{s i n A c o s A} = \frac{1}{c o s A} + \frac{1}{s i n A} = sec A + cosec A$ = RHS.
Q4 · 3 marks · Short answerHarder board level
If $3 sin A = 4 cos A$ , where $A$ is acute, find the value of $\frac{s e c A + cosec A}{s e c A - cosec A}$ and the value of $sin A cos A$ .
Hint
Turn the given equation into the value of a single ratio first.
Answer
$tan A = \frac{4}{3}$ . Dividing top and bottom by $cosec A$ and using $\frac{s e c A}{cosec A} = tan A$ : expression $= \frac{t a n A + 1}{t a n A - 1} = \frac{7/3}{1/3} = 7$ . Also $sec A = 1 + \frac{16}{9} = \frac{5}{3}$ , so $cos A = \frac{3}{5}$ , $sin A = tan A cos A = \frac{4}{5}$ and $sin A cos A = \frac{12}{25}$ . (Check: $sec A = \frac{5}{3}$ , $cosec A = \frac{5}{4}$ give $\frac{35/12}{5/12} = 7$ .)
Q5 · 2 marks · Short answerHOTS / competency
Sana claims that for some acute angle $A$ , $sin A = 0.8$ and $tan A = 0.6$ . Without finding $A$ , show that her claim is impossible.
Hint
Compare tan A with sin A using $tan A = \frac{s i n A}{c o s A}$ and $0 < cos A < 1$ .
Answer
$tan A = \frac{s i n A}{c o s A}$ and $0 < cos A < 1$ , so $tan A > sin A$ for every acute angle. But $0.6 < 0.8$ , a contradiction. (Directly: $cos A = 0.6$ , so $tan A = \frac{0.8}{0.6} = \frac{4}{3}$ , not 0.6.)
Q6 · 1 mark · Assertion-reasonBoard level
Assertion (A): $tan A > sin A$ for every acute angle $A$ .
Reason (R): For every acute angle $A$ , $0 < cos A < 1$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Write tan A as a quotient and ask what dividing by a number between 0 and 1 does.
Answer
(a). R is true for acute angles. Since $tan A = \frac{s i n A}{c o s A}$ with $sin A > 0$ and $0 < cos A < 1$ , dividing by $cos A$ makes the value larger, so A is true and R explains it.
Q7 · 3 marks · Short answerHarder board level
Prove that $(sec A - cosec A) (1 + tan A + cot A) = tan A sec A - cot A cosec A$ , where $A$ is acute.
Hint
Write both sides in $sin A$ and $cos A$ . On the right, use $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ .
Answer
Let $s = sin A$ , $c = cos A$ . LHS $= \frac{s - c}{sc} \cdot \frac{sc + s ^{2} + c ^{2}}{sc} = \frac{( s - c ) ( 1 + sc )}{s ^{2} c ^{2}}$ . RHS $= \frac{s}{c ^{2}} - \frac{c}{s ^{2}} = \frac{s ^{3} - c ^{3}}{s ^{2} c ^{2}} = \frac{( s - c ) ( s ^{2} + sc + c ^{2} )}{s ^{2} c ^{2}} = \frac{( s - c ) ( 1 + sc )}{s ^{2} c ^{2}}$ . Both sides are equal, so the result is proved.
Q8 · 3 marks · Short answerHarder board level
If $sec A + tan A = 3$ , where $A$ is acute, find $sec A - tan A$ , and hence find $sec A$ , $tan A$ and $sin A$ .
Hint
Use $sec^{2} A - tan^{2} A = 1$ and factorise the left side.
Answer
$(sec A + tan A) (sec A - tan A) = sec^{2} A - tan^{2} A = 1$ , so $sec A - tan A = \frac{1}{3}$ . Adding: $2 sec A = \frac{10}{3}$ , so $sec A = \frac{5}{3}$ ; subtracting: $tan A = \frac{4}{3}$ . $sin A = \frac{t a n A}{s e c A} = \frac{4}{3} \cdot \frac{3}{5} = \frac{4}{5}$ . Check: $\frac{5}{3} + \frac{4}{3} = 3$ .
Q9 · 5 marks · Long answerHOTS / competency
If $tan θ + sin θ = m$ and $tan θ - sin θ = n$ , where $θ$ is acute:
(i) Prove that $(m^{2} - n^{2})^{2} = 16 mn$ .
(ii) Find $m$ and $n$ when $θ = 6 0^{\circ}$ and verify the result for these values.
Hint
Find $m + n$ and $m - n$ first. For $mn$ , write $tan^{2} θ - sin^{2} θ$ over $cos^{2} θ$ .
Answer
(i) $m^{2} - n^{2} = (m + n) (m - n) = 2 tan θ \cdot 2 sin θ = 4 tan θ sin θ$ , so $(m^{2} - n^{2})^{2} = 16 tan^{2} θ sin^{2} θ$ . Also $mn = tan^{2} θ - sin^{2} θ = \frac{s i n ^{2} θ - s i n ^{2} θ c o s ^{2} θ}{c o s ^{2} θ} = \frac{s i n ^{2} θ ( 1 - c o s ^{2} θ )}{c o s ^{2} θ} = tan^{2} θ sin^{2} θ$ . So $(m^{2} - n^{2})^{2} = 16 mn$ .
(ii) $m = 3 + \frac{3}{2} = \frac{3 3}{2}$ and $n = 3 - \frac{3}{2} = \frac{3}{2}$ . $m^{2} - n^{2} = \frac{27}{4} - \frac{3}{4} = 6$ , so $(m^{2} - n^{2})^{2} = 36$ ; $16 mn = 16 \times \frac{9}{4} = 36$ . Verified.
Q10 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a tent is held up by a vertical pole, and a straight rope 1.7 m long runs from the top of the pole to a peg on the level ground. The rope makes an acute angle $A$ with the ground, and $cos A = \frac{8}{17}$ .
(i) Find $sin A$ . (1 mark)
(ii) Find $tan A$ and $sec A$ , and the height of the pole. (1 mark)
(iii) Find the value of $(sec A + tan A) (1 - sin A)$ , and show using the relationships between the ratios why it must equal $cos A$ for every acute angle. (2 marks)
Hint
For (iii), write $sec A + tan A$ as one fraction over $cos A$ .
Answer
(i) $sin A = 1 - \frac{64}{289} = \frac{15}{17}$ . (1)
(ii) $tan A = \frac{15}{8}$ , $sec A = \frac{17}{8}$ ; height $= 1.7 \times \frac{15}{17} = 1.5$ m. (1)
(iii) $(\frac{17}{8} + \frac{15}{8}) (1 - \frac{15}{17}) = 4 \times \frac{2}{17} = \frac{8}{17}$ . In general, $sec A + tan A = \frac{1 + s i n A}{c o s A}$ , so the product is $\frac{( 1 + s i n A ) ( 1 - s i n A )}{c o s A} = \frac{c o s ^{2} A}{c o s A} = cos A$ , which is $\frac{8}{17}$ here. (2)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Reciprocal of $sin A$ ?Show answer
$cosec A$
2. $tan A$ in terms of sin and cos?Show answer
$\frac{s i n A}{c o s A}$
3.Simplify $cos A tan A$ .Show answer
$sin A$
4.As A goes from 0° to 90°, does cot A increase or decrease?Show answer
Decreases.
5. $\frac{s e c A}{cosec A}$ equals?Show answer
$tan A$
6.Which two ratios are not defined at 90°?Show answer
$tan 9 0^{\circ}$ and $sec 9 0^{\circ}$ , because $cos 9 0^{\circ} = 0$ .

Relationships Between Trigonometric Ratios: frequently asked questions

Why can every trigonometric ratio be written using only sin and cos?

Cosec and sec are reciprocals of sin and cos, and tan and cot are quotients of sin and cos. So all six ratios are built from these two. That is why converting to sin and cos is the standard first step in simplification.

How do I prove that tan A equals sin A divided by cos A?

In a right triangle, tan A is the opposite side over the adjacent side. Divide both by the hypotenuse. The top becomes sin A and the bottom becomes cos A, so tan A equals sin A over cos A.

Is tan A always greater than sin A?

For an acute angle, yes. Tan A is sin A divided by cos A, and cos A lies between 0 and 1, so dividing by it makes the value larger.

Are complementary angle relations like sin of 90 minus A part of this topic?

No. Trigonometric ratios of complementary angles are not in the 2026-27 syllabus.