Trigonometry | ICSE Class 9 Maths Notes

This note covers right-angled triangles, trigonometric ratios and their reciprocals, standard-angle values, evaluation of expressions, finding unknown sides and angles in one right-angled triangle, and complementary-angle relationships.

What does trigonometry tell us about a right-angled triangle?

Trigonometry studies relationships between the sides and angles of a triangle. Here, the starting point is a right-angled triangle, which has one angle of 90°. The symbol ° means degrees, the unit of angle used throughout this note.

An acute angle is greater than 0° and less than 90°. The two other angles of a right-angled triangle are acute. They add to 90° because the three angles of a triangle add to 180°.

How do we name the sides?

Take triangle ABC, where A, B and C name its vertices, or corners. Let the right angle be at B. The notation AB means the side joining A and B, or its length when used in a calculation.

The hypotenuse is the longest side, opposite the right angle. Here it is AC. Choose angle A as the reference angle, meaning the acute angle from which the other side names are decided.

• Opposite side: BC faces angle A and does not form an arm of that angle.
• Adjacent side: AB touches angle A and is the side beside it other than the hypotenuse.
• Hypotenuse: AC remains opposite the right angle, whichever acute angle is selected.

Triangle ABC has A at the lower left, B at the lower right and C above B. A square marks the right angle at B. AC is labelled hypotenuse, BC opposite to angle A, and AB adjacent to angle A.

See Fig. 8.4 in your NCERT textbook

What changes when the reference angle changes?

Relative to angle C, AB becomes opposite and BC becomes adjacent. AC is still the hypotenuse. Thus, opposite and adjacent describe relationships to a chosen angle; they are not permanent names attached to particular sides.

Before calculating, mark the right angle and then the reference angle. Identify the hypotenuse first. This prevents the sloping appearance or orientation of a drawing from determining which side is placed in a ratio.

How are the six trigonometric ratios defined?

A ratio compares two quantities by division. A trigonometric ratio compares two side lengths of a right-angled triangle with respect to a chosen acute angle. Use the same length unit for both sides before dividing.

For triangle ABC right-angled at B, use angle A. The sine, cosine and tangent of A are written sin A, cos A and tan A. Each abbreviation belongs with its angle; sin A is not a multiplication of “sin” by A.

The final three rows define cosecant, secant and cotangent. A slash means division: BC/AC means the length BC divided by the length AC. These ratios have no length unit because the matching units cancel.

Property: A fixed angle has fixed ratios

When the acute angle remains the same, its trigonometric ratios do not change with the size of the triangle. Increasing the side lengths in the same proportion changes the triangle's size but leaves each corresponding side ratio unchanged.

Consequently, a ratio gives information about relative lengths. Knowing a ratio alone does not determine a particular triangle's side lengths. An actual side length provides the scale needed to calculate other lengths.

For an acute angle, sine and cosine are positive and less than 1, since their top numbers are lengths shorter than the hypotenuse. Tangent need not be less than 1: its value depends on which of the two shorter sides is longer.

How do reciprocal and quotient relationships simplify calculations?

The reciprocal of a non-zero number is 1 divided by that number. For a fraction, interchange its top and bottom numbers. The top number is its numerator; the bottom number, by which division occurs, is its denominator.

Result: Reciprocal pairs

For an acute angle A, the definitions give cosec A = 1/sin A, sec A = 1/cos A and cot A = 1/tan A. Thus, once sine, cosine and tangent are known, their three reciprocals follow directly.

The same pairs can be read in reverse: sin A = 1/cosec A, cos A = 1/sec A and tan A = 1/cot A. A reciprocal relationship can be used only when the quantities involved are defined and the divisor is non-zero.

Result: Tangent and cotangent as quotients

A quotient is the result of division. Since sin A = BC/AC and cos A = AB/AC, dividing them cancels the common hypotenuse. Therefore tan A = sin A/cos A. Similarly, cot A = cos A/sin A.

Worked example 1. For acute angle A, tan A = 4/3. Find the other five trigonometric ratios.

Answer: In triangle ABC right-angled at B, let BC = 4k and AB = 3k, where k is a positive common multiplier measured in the chosen length unit. This multiplier sets the scale of the triangle.

Pythagoras' theorem states that the square of the hypotenuse equals the sum of the squares of the other sides. The superscript ² means squaring, or multiplying a quantity by itself. Thus AC² = (4k)² + (3k)² = 25k², giving AC = 5k.

Hence sin A = 4/5, cos A = 3/5, cosec A = 5/4, sec A = 5/3 and cot A = 3/4.

This example also shows why tan A can exceed 1: the opposite side is longer than the adjacent side. Both sine and cosine still remain below 1.

How can Pythagoras' theorem supply a missing side?

To calculate a trigonometric ratio, its two side lengths must be available. If two sides of a right-angled triangle are known, Pythagoras' theorem supplies the third. Identify the hypotenuse before deciding whether to add or subtract the known squares.

If the hypotenuse is missing, add the squares of the two shorter sides. If a shorter side is missing, subtract the square of the known shorter side from the hypotenuse's square. Take the positive square root because a side length is positive.

The symbol √ denotes the non-negative square root: √400 = 20 because 20² = 400. When the square root is not a whole number, retain its exact form during calculation instead of introducing an early rounded value.

Worked example 2. Triangle ACB is right-angled at C. AB = 29 units and BC = 21 units. Let θ, pronounced theta, denote angle ABC, the angle at B. Find AC and evaluate cos² θ − sin² θ.

Answer: AB is the hypotenuse. AC² = AB² − BC² = 29² − 21² = 841 − 441 = 400. Therefore AC = 20 units.

Relative to θ, AC is opposite and BC adjacent. Thus sin θ = 20/29 and cos θ = 21/29. Here cos² θ means (cos θ)², and sin² θ means (sin θ)².

Substitution gives cos² θ − sin² θ = (21/29)² − (20/29)² = (441 − 400)/841 = 41/841.

How can a difference between sides be used?

Worked example 3. Triangle OPQ is right-angled at P. OP = 7 cm and OQ − PQ = 1 cm, where cm means centimetres. Find sin Q and cos Q.

Answer: OQ is the hypotenuse. Let x be the numerical length of PQ in centimetres. Then OQ has length x + 1 centimetres. Pythagoras gives (x + 1)² = 7² + x².

Expanding gives x² + 2x + 1 = 49 + x², so 2x = 48 and x = 24. Hence PQ = 24 cm and OQ = 25 cm.

At angle Q, OP is opposite and PQ adjacent. Therefore sin Q = 7/25 and cos Q = 24/25.

In both examples, the right angle determines the hypotenuse, while the requested acute angle determines opposite and adjacent. Keep these two decisions separate when labelling the triangle.

Why are the ratios of 45° especially simple?

An isosceles triangle has two equal sides. In a right-angled triangle with one acute angle of 45°, the other acute angle is also 45°. Equal angles have equal opposite sides, so the two sides meeting at the right angle are equal.

Take triangle ABC right-angled at B, with angles A and C both 45°. Let a be the positive common length of AB and BC. Pythagoras gives AC² = a² + a² = 2a², so AC = a√2.

A and B lie along the horizontal base, and C is above B. A square marks the right angle at B. An angle arc appears at A, and AC joins the two acute-angle vertices.

See Fig. 8.14 in your NCERT textbook

How do the equal sides determine all six ratios?

For angle A, opposite and adjacent both have length a. Dividing by the hypotenuse a√2 gives sin 45° = cos 45° = 1/√2. Dividing opposite by adjacent gives tan 45° = 1.

Taking the reciprocals gives cosec 45° = √2, sec 45° = √2 and cot 45° = 1. These are exact values: no measured decimal approximation of the triangle is needed.

Worked example 4. Triangle ABC is right-angled at B and tan A = 1. Verify that 2 sin A cos A = 1.

Answer: tan A = BC/AB = 1 means BC = AB. Write their common length as k, a positive number. Then AC² = k² + k², so AC = k√2.

Therefore sin A = 1/√2 and cos A = 1/√2. The expression 2 sin A cos A means twice the product of these ratios.

Thus 2 sin A cos A = 2 × (1/√2) × (1/√2) = 2/2 = 1, as required.

This conclusion uses the condition tan A = 1. The calculation verifies the expression for the stated triangle; it does not establish that the same product equals 1 for every acute angle.

How are the ratios of 30° and 60° obtained?

An equilateral triangle has three equal sides and three angles of 60°. Drawing a perpendicular from a vertex to the opposite side divides it into two matching right-angled triangles. Perpendicular lines meet at a right angle.

In equilateral triangle ABC, draw AD perpendicular to BC, with D on BC. The two right triangles ABD and ACD are congruent, meaning identical in shape and size. Consequently BD = DC, and AD divides angle A into two angles of 30°.

A is above the base BC, with D on that base beneath A. AD meets BC perpendicularly at D. The angle between AB and AD is marked 30°, and the angle at B is marked 60°.

See Fig. 8.15 in your NCERT textbook

What are the side lengths in the right triangle?

Let AB = 2a, where a is positive. Since BC also equals 2a and D divides BC equally, BD = a. Pythagoras in triangle ABD gives AD² = (2a)² − a² = 3a², so AD = a√3.

For the 30° angle at A, opposite is BD, adjacent is AD and hypotenuse is AB. Thus sin 30° = a/(2a) = 1/2, cos 30° = a√3/(2a) = √3/2, and tan 30° = a/(a√3) = 1/√3.

The reciprocals are cosec 30° = 2, sec 30° = 2/√3 and cot 30° = √3. The common factor a cancels, so these values do not depend on the size of the original triangle.

What changes for the 60° angle?

For the angle at B, AD becomes opposite and BD becomes adjacent. AB remains the hypotenuse. Therefore sin 60° = √3/2, cos 60° = 1/2 and tan 60° = √3.

The remaining values follow by reciprocation: cosec 60° = 2/√3, sec 60° = 2 and cot 60° = 1/√3. Comparing the 30° and 60° results shows how exchanging the two acute angles exchanges opposite and adjacent sides.

What are the standard-angle values, including 0° and 90°?

The standard angles used here are 0°, 30°, 45°, 60° and 90°. The acute-angle values come from right triangles. Values at 0° and 90° extend the ratios beyond the ordinary acute-angle construction.

How are the endpoint values understood?

When angle A in a right triangle becomes very close to 0°, the opposite side becomes very close to zero and the hypotenuse is nearly the same length as the adjacent side. Accordingly, sine is very close to 0 and cosine very close to 1.

At the endpoint, we define sin 0° = 0 and cos 0° = 1. When A becomes very close to 90°, sine becomes very close to 1 and cosine very close to 0. We define sin 90° = 1 and cos 90° = 0.

These statements distinguish being very close to an endpoint from the value defined at that endpoint. A triangle used for the acute-angle definitions has positive side lengths; it does not contain both a right angle and a zero angle.

Why are some entries not defined?

Not defined means that the ratio has no numerical value under these definitions. At 90°, tangent and secant would involve division by cos 90°, which is zero. At 0°, cosecant and cotangent would involve division by sin 0°, also zero.

Note: A zero numerator with a non-zero denominator gives zero. A zero denominator does not give a valid quotient. Keep “0” and “Not defined” distinct when using the table.

As A increases from 0° to 90°, sin A increases from 0 to 1, while cos A decreases from 1 to 0. Keep this angle range attached to the statement.

How should expressions containing standard angles be evaluated?

To evaluate an expression is to calculate its value. Replace each trigonometric ratio by its exact standard-angle value. Then carry out the indicated squares, products, divisions, additions and subtractions with the brackets preserved.

Read the notation carefully. The expression sin² A means the square of sin A, whereas sin 2A means the sine of twice angle A. Also, sin A cos A denotes multiplication of two ratio values.

What order keeps substitution clear?

1. Identify every angle and ratio in the original expression before selecting table entries.
2. Check that each required ratio and each denominator in the complete expression is defined and non-zero where necessary.
3. Substitute exact values, placing fractions inside brackets before squaring or multiplying them.
4. Simplify the resulting arithmetic and retain an exact fraction or square-root form when appropriate.

Worked example 5. Evaluate sin 60° cos 30° + sin 30° cos 60°.

Answer: sin 60° and cos 30° both equal √3/2. Meanwhile sin 30° and cos 60° both equal 1/2.

The expression becomes (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1.

Each pair is multiplied before the two products are added. Writing these products separately makes the arithmetic visible and avoids treating adjacent ratio names as one unfamiliar operation.

Worked example 6. Evaluate 2 tan² 45° + cos² 30° − sin² 60°.

Answer: tan 45° = 1, while cos 30° = sin 60° = √3/2. Therefore the expression equals 2 × 1² + (√3/2)² − (√3/2)².

This gives 2 + 3/4 − 3/4 = 2. The two equal squared terms cancel because one is added and the other subtracted.

The coefficient 2 multiplies the squared tangent value. It does not change the angle 45° into another angle. Substitution first, followed by arithmetic, keeps this distinction clear.

How can one right-angled triangle give unknown lengths and angles?

A two-dimensional problem describes a situation in a plane. When it is represented by one right-angled triangle, a known side and a known acute angle can determine the remaining sides. Two known sides can also determine an acute angle through a ratio.

Choose the ratio containing the information given and the quantity required. Sine connects opposite and hypotenuse, cosine connects adjacent and hypotenuse, and tangent connects opposite and adjacent. This choice avoids introducing an unnecessary unknown side.

Worked example 7. Triangle ABC is right-angled at B. AB = 5 cm and angle ACB = 30°. Find BC and AC.

Answer: Relative to angle C, AB is opposite, BC adjacent and AC the hypotenuse. Therefore tan 30° = AB/BC gives 1/√3 = 5/BC.

Multiplying both sides by BC√3 gives BC = 5√3 cm. To find AC, use sin 30° = AB/AC, so 1/2 = 5/AC and AC = 10 cm.

As a check, AB² + BC² = 25 + 75 = 100 cm², equal to AC². The symbol cm² means square centimetres.

How can a known ratio identify an angle?

Worked example 8. Triangle PQR is right-angled at Q. PQ = 3 cm and PR = 6 cm. Find angles QPR and PRQ.

Answer: Angle PRQ is at R, and PQ is opposite it. PR is the hypotenuse. Hence sin R = PQ/PR = 3/6 = 1/2.

For an acute angle, this is the sine value of 30°, so angle PRQ = 30°. The other acute angle is angle QPR = 90° − 30° = 60°.

In a three-letter angle name, the middle letter identifies the vertex. Naming the requested angle fully helps distinguish an answer at P from an answer at R.

Finish length calculations with units, but give ratio values without length units. When solving for an angle, include the degree symbol. Check that the hypotenuse is longer than either shorter side and that the acute angles together make 90°.

How do complementary angles connect the six ratios?

Complementary angles are two angles whose sum is 90°. Thus the acute angles A and C in triangle ABC right-angled at B are complementary, with C = 90° − A. Both angles are measured in degrees.

Changing the reference angle from A to C exchanges opposite and adjacent sides but preserves the hypotenuse. Sine for one angle therefore equals cosine for the other. The same side exchange connects tangent with cotangent and secant with cosecant.

Result: Complementary-angle relationships

These relationships apply directly to acute complementary angles. At endpoints, use an equality only where both ratios are defined. For instance, sin 0° = cos 90° = 0, but neither tan 90° nor cot 0° has a defined value.

How does switching angles work numerically?

Worked example 9. Triangle ABC is right-angled at B, with AB = 24 cm and BC = 7 cm. Find sin A, cos A, sin C and cos C.

Answer: AC² = 24² + 7² = 576 + 49 = 625, so AC = 25 cm. At A, BC is opposite and AB adjacent, giving sin A = 7/25 and cos A = 24/25.

At C, AB is opposite and BC adjacent. Hence sin C = 24/25 and cos C = 7/25. Therefore sin A = cos C and cos A = sin C.

Direct applications also follow from the standard-angle table: sin 30° = cos 60°, tan 30° = cot 60°, and sec 30° = cosec 60°. Each equality compares the appropriate paired ratios at angles that total 90°.

Note: A complementary-angle relationship changes both the angle and the ratio name. Sine changes to cosine, tangent to cotangent, and secant to cosecant, or the reverse in each pair.

Glossary

• Trigonometry — The study of relationships between the sides and angles of triangles.
• Right-angled triangle — A triangle containing one right angle, with its other two angles acute.
• Acute angle — An angle greater than zero degrees and less than ninety degrees.
• Hypotenuse — The longest side of a right-angled triangle, opposite its right angle.
• Opposite side — The side facing the chosen acute angle in a right-angled triangle.
• Adjacent side — The side beside the chosen acute angle, other than the hypotenuse.
• Sine — The ratio of the opposite side to the hypotenuse for an acute angle.
• Cosine — The ratio of the adjacent side to the hypotenuse for an acute angle.
• Tangent — The ratio of the opposite side to the adjacent side for an acute angle.
• Reciprocal — The number obtained by dividing one by a specified non-zero number.
• Cosecant — The reciprocal of sine, equal to hypotenuse divided by opposite side for an acute angle.
• Secant — The reciprocal of cosine, equal to hypotenuse divided by adjacent side for an acute angle.
• Cotangent — The reciprocal of tangent, equal to adjacent side divided by opposite side for an acute angle.
• Complementary angles — Two angles whose measures add together to give exactly ninety degrees.
• Standard angles — The angles 0°, 30°, 45°, 60° and 90°, whose trigonometric values are tabulated here.

Common errors and misconceptions

• Misconception: The opposite side stays the same whichever acute angle is chosen. Correct: Opposite and adjacent exchange roles when the reference angle changes to the other acute angle. The hypotenuse remains unchanged.
• Misconception: sin A means “sin” multiplied by A. Correct: The complete notation means the sine of angle A. The abbreviation and its angle must be read together.
• Misconception: Tangent must be less than 1 because it is a ratio. Correct: Tangent compares opposite with adjacent. If opposite is longer, tangent exceeds 1, as tan 60° = √3 shows.
• Misconception: cosec A is another name for cos A. Correct: Cosecant is the reciprocal of sine. Cosine is the adjacent side divided by the hypotenuse.
• Misconception: tan 90° and cot 0° equal zero. Correct: Both are not defined because their denominator becomes zero. The defined zero values are tan 0° and cot 90°.
• Misconception: sin² A and sin 2A mean the same thing. Correct: The first squares the sine value; the second takes the sine of twice the angle.
• Misconception: sin A = cos A for every acute angle. Correct: Sine equals the cosine of the complementary angle. Their values at the same angle agree at 45° in the acute-angle range.
• Misconception: Ratio values carry centimetres when the sides are measured in centimetres. Correct: Matching length units cancel in the division. Actual calculated side lengths still require their units.

Exam-style questions with model answers

Q1. Triangle ABC is right-angled at B. Define sin A and cos A using its named sides, identifying the hypotenuse. [2 marks]

1. AC is the hypotenuse, opposite the right angle. Since BC is opposite angle A, sin A = BC/AC.
2. AB is adjacent to angle A and is not the hypotenuse. Therefore cos A = AB/AC.

Q2. Triangle ABC is right-angled at B, AB = 5 cm and angle ACB = 30°. Find BC and AC. [3 marks]

1. Using angle C, AB is the opposite side and BC is the adjacent side. Therefore tan 30° = AB/BC, which gives 1/√3 = 5/BC.
2. Multiplying this equation by BC√3 gives BC = 5√3 cm. This is the length of the side adjacent to the given angle.
3. For the hypotenuse, sin 30° = AB/AC gives 1/2 = 5/AC. Multiplication by 2AC gives AC = 10 cm.

Q3. Evaluate 2 tan² 45° + cos² 30° − sin² 60°, showing the substitutions and squaring. [4 marks]

1. The standard value tan 45° = 1 gives the first term as 2 tan² 45° = 2 × 1² = 2.
2. Since cos 30° = √3/2, the second term is cos² 30° = (√3/2)² = 3/4.
3. Since sin 60° = √3/2, the term being subtracted is sin² 60° = (√3/2)² = 3/4.
4. Combining the three terms gives 2 + 3/4 − 3/4 = 2. The equal fractional terms cancel.

Q4. Angle A is acute and tan A = 4/3. Find the other five trigonometric ratios, using a right-angled triangle. [5 marks]

1. Choose triangle ABC right-angled at B, with BC = 4k and AB = 3k, where k is a positive scale factor. Pythagoras gives AC = 5k. Therefore sin A = BC/AC = 4/5.
2. The adjacent side is AB and the hypotenuse is AC, so cos A = AB/AC = 3/5.
3. Cosecant is the reciprocal of sine. Hence cosec A = AC/BC = 5/4.
4. Secant is the reciprocal of cosine. Hence sec A = AC/AB = 5/3.
5. Cotangent is adjacent divided by opposite, the reciprocal of tangent. Hence cot A = AB/BC = 3/4.

Q5. Triangle PQR is right-angled at Q, PQ = 3 cm and PR = 6 cm. Calculate angles PRQ and QPR. [3 marks]

1. PR is the hypotenuse because it faces the right angle at Q. Relative to angle R, PQ is opposite, so sin R = PQ/PR = 3/6 = 1/2.
2. The acute angle with sine equal to 1/2 is 30°. Therefore angle PRQ, whose vertex is R, equals 30°.
3. The two acute angles of the right triangle total 90°. Therefore angle QPR = 90° − 30° = 60°.

Q6. Triangle ABC is right-angled at B, AB = 24 cm and BC = 7 cm. Calculate sin A, cos A, sin C and cos C, and show the complementary-angle equalities. [4 marks]

1. Pythagoras gives AC² = AB² + BC² = 24² + 7² = 625. Taking the positive square root gives the hypotenuse AC = 25 cm.
2. For angle A, BC is opposite and AB adjacent. Thus sin A = 7/25 and cos A = 24/25.
3. For angle C, AB is opposite and BC adjacent. Thus sin C = 24/25 and cos C = 7/25.
4. Comparing these values gives sin A = cos C and cos A = sin C, consistent with A and C being complementary.

Q7. State the values of tan 0° and cot 90°, and explain why tan 90° and cot 0° are not defined. [3 marks]

1. Using tan A = sin A/cos A, tan 0° = 0/1 = 0. Using cot A = cos A/sin A, cot 90° = 0/1 = 0.
2. At 90°, sin 90° = 1 and cos 90° = 0. Thus tangent would require the quotient 1/0, so tan 90° is not defined.
3. At 0°, cos 0° = 1 and sin 0° = 0. Cotangent would also require division by zero, so cot 0° is not defined.

Q8. Triangle OPQ is right-angled at P, OP = 7 cm and OQ − PQ = 1 cm. Find PQ, OQ, sin Q and cos Q. [5 marks]

1. OQ is the hypotenuse. Let x be the numerical length of PQ in centimetres. The given difference makes the numerical length of OQ equal to x + 1.
2. Apply Pythagoras to obtain (x + 1)² = 7² + x². Expanding both sides gives x² + 2x + 1 = 49 + x².
3. Cancel x² to obtain 2x = 48, so x = 24. Therefore PQ = 24 cm and OQ = 25 cm.
4. For angle Q, OP is opposite and OQ is the hypotenuse. Hence sin Q = OP/OQ = 7/25.
5. PQ is adjacent to angle Q. Hence cos Q = PQ/OQ = 24/25.

Key takeaways

• Choose the reference angle before naming opposite and adjacent sides; the hypotenuse remains opposite the right angle.
• Sine compares opposite with hypotenuse, cosine compares adjacent with hypotenuse, and tangent compares opposite with adjacent.
• Cosecant, secant and cotangent are the reciprocals of sine, cosine and tangent respectively, wherever the required division is defined.
• Use Pythagoras' theorem to find a missing side, then substitute side lengths into the required ratio definitions.
• Know the exact values at 0°, 30°, 45°, 60° and 90°, keeping undefined entries distinct from zero.
• For standard-angle expressions, substitute exact values before squaring or multiplying, and preserve the original brackets and operations.
• Complementary angles add to 90°; exchanging them pairs sine with cosine, tangent with cotangent, and secant with cosecant.
• One known side and one acute angle can determine the remaining sides of a right-angled triangle.

Test yourself

In triangle ABC right-angled at B, which side is opposite angle C?

AB is opposite angle C. BC is adjacent to C, while AC remains the hypotenuse.

Why does changing a triangle's size leave the ratios of a fixed angle unchanged?

The corresponding side lengths change in the same proportion, so that common factor cancels when one length is divided by another.

What are the reciprocal partners of sine, cosine and tangent?

They are cosecant, secant and cotangent respectively. Each reciprocal relationship requires a defined value and division by a non-zero quantity.

Why is tan 45° equal to 1?

A right triangle with acute angles of 45° has equal opposite and adjacent sides, so their quotient is 1.

What does sin² A mean?

It means (sin A)², the square of the sine value. It does not mean the sine of twice the angle.

Which four standard-angle entries are not defined?

They are cosec 0°, cot 0°, tan 90° and sec 90°. Each would require division by zero.

How does cos A change as A increases from 0° to 90°?

Cosine decreases from 1 to 0 over this angle range, while sine increases from 0 to 1.

Express sec A using the complementary angle for acute A.

sec A = cosec (90° − A). The angle changes to its complement and secant changes to cosecant.