Introduction To Trigonometry

Welcome dear students! Today we are going to learn about Introduction To Trigonometry from Class 10 Maths.

There is perhaps nothing which so occupies the middle position of mathematics as trigonometry, as noted by J.F. Herbart in 1890. You have already studied about triangles, and in particular, right triangles, in your earlier classes. Let us take some examples from our surroundings where right triangles can be imagined to be formed. For instance, suppose the students of a school are visiting Qutub Minar. Now, if a student is looking at the top of the Minar, a right triangle can be imagined to be made. Can the student find out the height of the Minar, without actually measuring it? Suppose a girl is sitting on the balcony of her house located on the bank of a river. She is looking down at a flower pot placed on a stair of a temple situated nearby on the other bank of the river. A right triangle is imagined to be made in this situation. If you know the height at which the person is sitting, can you find the width of the river? Suppose a hot air balloon is flying in the air. A girl happens to spot the balloon in the sky and runs to her mother to tell her about it. Her mother rushes out of the house to look at the balloon. Now when the girl had spotted the balloon initially it was at point A. When both the mother and daughter came out to see it, it had already travelled to another point B. Can you find the altitude of B from the ground? In all the situations given above, the distances or heights can be found by using some mathematical techniques, which come under a branch of mathematics called ‘trigonometry’. The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact, trigonometry is the study of relationships between the sides and angles of a triangle. The earliest known work on trigonometry was recorded in Egypt and Babylon. Early astronomers used it to find out the distances of the stars and planets from the Earth. Even today, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts. In this chapter, we will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle. We will restrict our discussion to acute angles only. However, these ratios can be extended to other angles also. We will also define the trigonometric ratios for angles of measure 0° and 90°. We will calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities. [CHECKPOINT]

Let us move to Section 8.2, Trigonometric Ratios. Let us take a right triangle ABC, right-angled at B. Here, angle A is an acute angle. Note the position of the side BC with respect to angle A. It faces angle A. We call it the side opposite to angle A. AC is the hypotenuse of the right triangle and the side AB is a part of angle A. So, we call it the side adjacent to angle A. Note that the position of sides change when you consider angle C in place of A. You have studied the concept of ‘ratio’ in your earlier classes. We now define certain ratios involving the sides of a right triangle, and call them trigonometric ratios. The trigonometric ratios of the angle A in right triangle ABC are defined as follows: sine of angle A = side opposite to angle A / hypotenuse = BC / AC cosine of angle A = side adjacent to angle A / hypotenuse = AB / AC tangent of angle A = side opposite to angle A / side adjacent to angle A = BC / AB cosecant of angle A = 1 / sine of angle A = hypotenuse / side opposite to angle A = AC / BC secant of angle A = 1 / cosine of angle A = hypotenuse / side adjacent to angle A = AC / AB cotangent of angle A = 1 / tangent of angle A = side adjacent to angle A / side opposite to angle A = AB / BC The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A and cot A respectively. Note that the ratios cosec A, sec A and cot A are respectively, the reciprocals of the ratios sin A, cos A and tan A. Also, observe that tan A = sin A / cos A and cot A = cos A / sin A. So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides. Why don’t you try to define the trigonometric ratios for angle C in the right triangle? [CHECKPOINT]

The first use of the idea of ‘sine’ in the way we use it today was in the work Aryabhatiyam by Aryabhata, in A.D. 500. Aryabhata used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course. When the Aryabhatiyam was translated into Arabic, the word jiva was retained as it is. The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin. Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe. An English Professor of astronomy Edmund Gunter (1581–1626), first used the abbreviated notation ‘sin’. The origin of the terms ‘cosine’ and ‘tangent’ was much later. The cosine function arose from the need to compute the sine of the complementary angle. Aryabhatta called it kotijya. The name cosinus originated with Edmund Gunter. In 1674, the English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’. Aryabhata lived during C.E. 476 – 550. Remark: Note that the symbol sin A is used as an abbreviation for ‘the sine of the angle A’. sin A is not the product of ‘sin’ and A. ‘sin’ separated from A has no meaning. Similarly, cos A is not the product of ‘cos’ and A. Similar interpretations follow for other trigonometric ratios also. Note that cosec A = (sin A)⁻¹, which is not equal to sin⁻¹ A. sin⁻¹ A is called sine inverse A and has a different meaning, which will be discussed in higher classes. Similar conventions hold for the other trigonometric ratios as well. Sometimes, the Greek letter θ (theta) is also used to denote an angle. We have defined six trigonometric ratios of an acute angle. If we know any one of the ratios, can we obtain the other ratios? Let us see. If in a right triangle ABC, sin A = 1/3, then this means that BC / AC = 1/3, i.e., the lengths of the sides BC and AC of the triangle ABC are in the ratio 1 : 3. So if BC is equal to k, then AC will be 3k, where k is any positive number. To determine other trigonometric ratios for the angle A, we need to find the length of the third side AB. Do you remember the Pythagoras theorem? Let us use it to determine the required length AB. AB² = AC² – BC² = (3k)² – (k)² = 8k² = (2√2 k)². Therefore, AB = 2√2 k. (Why is AB not – 2√2 k?) Now, cos A = AB / AC = 2√2 / 3. Similarly, you can obtain the other trigonometric ratios of the angle A. Remark: Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or, in particular, equal to 1). [CHECKPOINT]

Now, if we take a point P on the hypotenuse AC or a point Q on AC extended, of the right triangle ABC and draw PM perpendicular to AB and QN perpendicular to AB extended (see Fig. 8.6), how will the trigonometric ratios of angle A in triangle PAM differ from those of angle A in triangle CAB or from those of angle A in triangle QAN? To answer this, first look at these triangles. Is triangle PAM similar to triangle CAB? From Chapter 6, recall the AA similarity criterion. Using the criterion, you will see that the triangles PAM and CAB are similar. Therefore, by the property of similar triangles, the corresponding sides of the triangles are proportional. So, we have AM / AB = MP / BC = AP / AC. From this, we find MP / AP = BC / AC = sin A. Similarly, AM / AP = AB / AC = cos A, and MP / AM = BC / AB = tan A, and so on. This shows that the trigonometric ratios of angle A in triangle PAM do not differ from those of angle A in triangle CAB. In the same way, you should check that the value of sin A (and also of other trigonometric ratios) remains the same in triangle QAN also. From our observations, it is now clear that the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.

Let us consider some examples. Example 1: Given tan A = 4/3, find the other trigonometric ratios of the angle A. Solution: Let us first draw a right triangle ABC. Now, we know that tan A = BC / AB = 4/3. Therefore, if BC = 4k, then AB = 3k, where k is a positive number. Now, by using the Pythagoras Theorem, we have AC² = AB² + BC² = (3k)² + (4k)² = 9k² + 16k² = 25k². So, AC = 5k. Now, we can write all the trigonometric ratios using their definitions. sin A = BC / AC = 4/5. cos A = AB / AC = 3/5. Therefore, cot A = 1 / tan A = 3/4, cosec A = 1 / sin A = 5/4, and sec A = 1 / cos A = 5/3.

Example 2: If angle B and angle Q are acute angles such that sin B = sin Q, then prove that angle B = angle Q. Solution: Let us consider two right triangles ABC and PQR where sin B = sin Q. We have sin B = AC / AB and sin Q = PR / PQ. Then AC / AB = PR / PQ. Therefore, AC / PR = AB / PQ = k, say. Now, using Pythagoras theorem, BC = √(AB² – AC²) and QR = √(PQ² – PR²). So, BC / QR = √(AB² – AC²) / √(PQ² – PR²) = √(k²PQ² – k²PR²) / √(PQ² – PR²) = k√(PQ² – PR²) / √(PQ² – PR²) = k. From these, we have AC / PR = AB / PQ = BC / QR. Then, by using the SSS similarity criterion, triangle ACB is similar to triangle PRQ and therefore, angle B = angle Q. [CHECKPOINT]

Example 3: Consider triangle ACB, right-angled at C, in which AB = 29 units, BC = 21 units and angle ABC = θ. Determine the values of (i) cos² θ + sin² θ, (ii) cos² θ – sin² θ. Solution: In triangle ACB, we have AC = √(AB² – BC²) = √(29² – 21²) = √((29 – 21)(29 + 21)) = √(8 × 50) = √400 = 20 units. So, sin θ = AC / AB = 20/29, cos θ = BC / AB = 21/29. Now, (i) cos² θ + sin² θ = (21/29)² + (20/29)² = (441 + 400)/841 = 841/841 = 1, and (ii) cos² θ – sin² θ = (21/29)² – (20/29)² = (441 – 400)/841 = 41/841.

Example 4: In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1. Solution: In triangle ABC, tan A = BC / AB = 1. i.e., BC = AB. Let AB = BC = k, where k is a positive number. Now, AC = √(AB² + BC²) = √(k² + k²) = √(2k²) = k√2. Therefore, sin A = BC / AC = 1/√2 and cos A = AB / AC = 1/√2. So, 2 sin A cos A = 2 × (1/√2) × (1/√2) = 2 × (1/2) = 1, which is the required value.

Example 5: In triangle OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm. Determine the values of sin Q and cos Q. Solution: In triangle OPQ, we have OQ² = OP² + PQ². i.e., (1 + PQ)² = OP² + PQ². i.e., 1 + PQ² + 2PQ = OP² + PQ². i.e., 1 + 2PQ = 7². i.e., 2PQ = 49 – 1 = 48, so PQ = 24 cm and OQ = 1 + PQ = 25 cm. So, sin Q = OP / OQ = 7/25 and cos Q = PQ / OQ = 24/25. [CHECKPOINT]

Now let us move to Exercise 8.1. Question 1: In triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C. Solution: First, find hypotenuse AC using Pythagoras theorem. AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625. So AC = 25 cm. For angle A, opposite is BC, adjacent is AB. So sin A = BC / AC = 7/25. cos A = AB / AC = 24/25. For angle C, opposite is AB, adjacent is BC. So sin C = AB / AC = 24/25. cos C = BC / AC = 7/25.

Question 2: In Fig. 8.13, find tan P – cot R. Solution: Figure 8.13 shows a right triangle PQR, right-angled at Q. The side PQ measures 12 units, the side QR measures 5 units, and the hypotenuse PR measures 13 units. In this triangle, tan P is the ratio of the side opposite to angle P to the side adjacent to angle P, which is QR / PQ = 5/12. Cot R is the ratio of the side adjacent to angle R to the side opposite to angle R, which is also QR / PQ = 5/12. Therefore, tan P – cot R = 5/12 – 5/12 = 0.

Question 3: If sin A = 3/4, calculate cos A and tan A. Solution: sin A = opposite / hypotenuse = 3/4, so let BC = 3k and AC = 4k. By Pythagoras theorem, AB² = AC² – BC² = (4k)² – (3k)² = 16k² – 9k² = 7k². So AB = k√7. cos A = AB / AC = √7/4. tan A = BC / AB = 3/√7.

Question 4: Given 15 cot A = 8, find sin A and sec A. Solution: cot A = 8/15. So adjacent = 8k, opposite = 15k. Hypotenuse² = (8k)² + (15k)² = 64k² + 225k² = 289k². So hypotenuse = 17k. sin A = opposite / hypotenuse = 15/17. sec A = hypotenuse / adjacent = 17/8.

Question 5: Given sec θ = 13/12, calculate all other trigonometric ratios. Solution: sec θ = hypotenuse / adjacent = 13/12, so let AC = 13k, AB = 12k. BC² = AC² – AB² = (13k)² – (12k)² = 169k² – 144k² = 25k². So BC = 5k. sin θ = BC / AC = 5/13, cos θ = AB / AC = 12/13, tan θ = BC / AB = 5/12, cosec θ = AC / BC = 13/5, cot θ = AB / BC = 12/5.

Question 6: If angle A and angle B are acute angles such that cos A = cos B, then show that angle A = angle B. Solution: In right triangles, cos A = adjacent / hypotenuse, and cos B = adjacent / hypotenuse. Since cos A = cos B, the ratios of corresponding sides are equal. By the same logic as Example 2, the triangles are similar by SSS criterion, and thus angle A = angle B. [CHECKPOINT]

Question 7: If cot θ = 7/8, evaluate: (i) (1 + sin θ)(1 – sin θ) / (1 + cos θ)(1 – cos θ), (ii) cot² θ. Solution: cot θ = 7/8, so adjacent is 7k, opposite is 8k. Hypotenuse² = (7k)² + (8k)² = 49k² + 64k² = 113k². So hypotenuse = k√113. sin θ = 8/√113, cos θ = 7/√113. The first expression simplifies to (1 – sin² θ) / (1 – cos² θ) = cos² θ / sin² θ = cot² θ. cot² θ = (7/8)² = 49/64. So both parts equal 49/64.

Question 8: If 3 cot A = 4, check whether (1 – tan² A) / (1 + tan² A) = cos² A – sin² A or not. Solution: cot A = 4/3. So tan A = 3/4. LHS = (1 – (3/4)²) / (1 + (3/4)²) = (1 – 9/16) / (1 + 9/16) = (7/16) / (25/16) = 7/25. For RHS, sides are 3, 4, 5. cos A = 4/5, sin A = 3/5. cos² A – sin² A = (16/25) – (9/25) = 7/25. LHS = RHS, so the statement is true.

Question 9: In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of: (i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C. Solution: tan A = 1/√3, so opposite is k, adjacent is k√3. Hypotenuse² = k² + (k√3)² = k² + 3k² = 4k², so hypotenuse is 2k. sin A = 1/2, cos A = √3/2. Since angle B = 90°, angle C = 60°. sin C = √3/2, cos C = 1/2. First expression: (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1. Second expression: (√3/2)(1/2) – (1/2)(√3/2) = √3/4 – √3/4 = 0.

Question 10: In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. Solution: Let QR = x. Then PR = 25 – x. By Pythagoras, PR² = PQ² + QR². So (25 – x)² = 5² + x². 625 – 50x + x² = 25 + x². 50x = 600, so x = 12. QR = 12, PR = 13. sin P = QR / PR = 12/13. cos P = PQ / PR = 5/13. tan P = QR / PQ = 12/5.

Question 11: State whether the following are true or false. Justify your answer. (i) The value of tan A is always less than 1. False, it can be greater than 1. (ii) sec A = 12/5 for some value of angle A. True, sec A can be greater than 1. (iii) cos A is the abbreviation used for the cosecant of angle A. False, it is cosine. (iv) cot A is the product of cot and A. False, it is a single symbol. (v) sin θ = 4/3 for some angle θ. False, sine cannot exceed 1. [CHECKPOINT]

Now let us move to Section 8.3, Trigonometric Ratios of Some Specific Angles. From geometry, you are already familiar with the construction of angles of 30°, 45°, 60° and 90°. In this section, we will find the values of the trigonometric ratios for these angles and, of course, for 0°. Trigonometric Ratios of 45°: In triangle ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., angle A = angle C = 45°. So, BC = AB. Suppose BC = AB = a. Then by Pythagoras Theorem, AC² = AB² + BC² = a² + a² = 2a². Therefore, AC = a√2. Using the definitions, sin 45° = BC / AC = 1/√2. cos 45° = AB / AC = 1/√2. tan 45° = BC / AB = 1. Also, cosec 45° = √2, sec 45° = √2, cot 45° = 1.

Trigonometric Ratios of 30° and 60°: Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore, angle A = angle B = angle C = 60°. Draw the perpendicular AD from A to the side BC. Now triangle ABD is congruent to triangle ACD by RHS congruence. Therefore, BD = DC and angle BAD = angle CAD. Now observe that: triangle ABD is a right triangle, right-angled at D with angle BAD = 30° and angle ABD = 60°. As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that AB = 2a. Then, BD = a. And AD² = AB² – BD² = (2a)² – (a)² = 3a². Therefore, AD = a√3. Now, we have: sin 30° = BD / AB = 1/2. cos 30° = AD / AB = √3/2. tan 30° = BD / AD = 1/√3. Also, cosec 30° = 2, sec 30° = 2/√3, cot 30° = √3. Similarly, sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3, cosec 60° = 2/√3, sec 60° = 2, cot 60° = 1/√3.

Trigonometric Ratios of 0° and 90°: Let us see what happens to the trigonometric ratios of angle A, if it is made smaller and smaller in the right triangle ABC, till it becomes zero. As angle A gets smaller and smaller, the length of the side BC decreases. The point C gets closer to point B, and finally when angle A becomes very close to 0°, AC becomes almost the same as AB. When angle A is very close to 0°, BC gets very close to 0 and so the value of sin A = BC / AC is very close to 0. Also, when angle A is very close to 0°, AC is nearly the same as AB and so the value of cos A = AB / AC is very close to 1. This helps us to see how we can define the values of sin A and cos A when A = 0°. We define: sin 0° = 0 and cos 0° = 1. Using these, we have: tan 0° = 0, cot 0° is not defined. sec 0° = 1 and cosec 0° is not defined. Now, let us see what happens to the trigonometric ratios of angle A, when it is made larger and larger in triangle ABC till it becomes 90°. As angle A gets larger and larger, angle C gets smaller and smaller. Therefore, as in the case above, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when angle A is very close to 90°, angle C becomes very close to 0° and the side AC almost coincides with side BC. When angle C is very close to 0°, angle A is very close to 90°, side AC is nearly the same as side BC, and so sin A is very close to 1. Also when angle A is very close to 90°, angle C is very close to 0°, and the side AB is nearly zero, so cos A is very close to 0. So, we define: sin 90° = 1 and cos 90° = 0. Now, why don’t you find the other trigonometric ratios of 90°? We shall now give the values of all the trigonometric ratios of 0°, 30°, 45°, 60° and 90° in Table 8.1, for ready reference. [CHECKPOINT]

Table 8.1 lists the following values: For angle A = 0°: sin A = 0, cos A = 1, tan A = 0, cosec A = Not defined, sec A = 1, cot A = Not defined. For angle A = 30°: sin A = 1/2, cos A = √3/2, tan A = 1/√3, cosec A = 2, sec A = 2/√3, cot A = √3. For angle A = 45°: sin A = 1/√2, cos A = 1/√2, tan A = 1, cosec A = √2, sec A = √2, cot A = 1. For angle A = 60°: sin A = √3/2, cos A = 1/2, tan A = √3, cosec A = 2/√3, sec A = 2, cot A = 1/√3. For angle A = 90°: sin A = 1, cos A = 0, tan A = Not defined, cosec A = 1, sec A = Not defined, cot A = 0. Remark: From the table above you can observe that as angle A increases from 0° to 90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0.

Let us illustrate the use of the values in the table above through some examples. Example 6: In triangle ABC, right-angled at B, AB = 5 cm and angle ACB = 30°. Determine the lengths of the sides BC and AC. Solution: To find the length of the side BC, we will choose the trigonometric ratio involving BC and the given side AB. Since BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore AB / BC = tan C. i.e., 5 / BC = tan 30° = 1/√3, which gives BC = 5√3 cm. To find the length of the side AC, we consider sin 30° = AB / AC. i.e., 1/2 = 5 / AC, i.e., AC = 10 cm. Note that alternatively we could have used Pythagoras theorem to determine the third side in the example above, i.e., AC = √(AB² + BC²) = √(25 + 75) = √100 = 10 cm.

Example 7: In triangle PQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine angle QPR and angle PRQ. Solution: Given PQ = 3 cm and PR = 6 cm. Therefore, PQ / PR = sin R or sin R = 3/6 = 1/2. So, angle PRQ = 30° and therefore, angle QPR = 60°. You may note that if one of the sides and any other part (either an acute angle or any side) of a right triangle is known, the remaining sides and angles of the triangle can be determined.

Example 8: If sin (A – B) = 1/2, cos (A + B) = 1/2, 0° < A + B ≤ 90°, A > B, find A and B. Solution: Since, sin (A – B) = 1/2, therefore, A – B = 30°. Also, since cos (A + B) = 1/2, therefore, A + B = 60°. Solving these two equations, we get: A = 45° and B = 15°. [CHECKPOINT]

Now let us move to Exercise 8.2. Question 1: Evaluate the following: (i) sin 60° cos 30° + sin 30° cos 60° = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1. (ii) 2 tan² 45° + cos² 30° – sin² 60° = 2(1)² + (√3/2)² – (√3/2)² = 2 + 3/4 – 3/4 = 2. (iii) cos 45° sec 30° + cosec 30° = (1/√2)(2/√3) + 2 = 2/√6 + 2 = (2√6)/6 + 2 = √6/3 + 2. (iv) (sin 30° + tan 45° – cosec 60°) / (sec 30° + cos 60° + cot 45°) = (1/2 + 1 – 2/√3) / (2/√3 + 1/2 + 1) = (3/2 – 2/√3) / (3/2 + 2/√3). Multiplying numerator and denominator by 2√3 gives (3√3 – 4) / (3√3 + 4). Rationalizing the denominator by multiplying numerator and denominator by (3√3 – 4) yields (27 + 16 – 24√3) / (27 – 16) = (43 – 24√3) / 11. (v) (5 cos² 60° + 4 sec² 30° – tan² 45°) / (sin² 30° + cos² 30°) = (5(1/2)² + 4(2/√3)² – 1²) / ((1/2)² + (√3/2)²) = (5/4 + 16/3 – 1) / (1/4 + 3/4) = (15/12 + 64/12 – 12/12) / 1 = 67/12.

Question 2: Choose the correct option and justify your choice: (i) 2 tan 30° / (1 + tan² 30°) = 2(1/√3) / (1 + 1/3) = (2/√3) / (4/3) = 6/(4√3) = √3/2 = sin 60°. Option (A). (ii) (1 – tan² 45°) / (1 + tan² 45°) = (1 – 1) / (1 + 1) = 0/2 = 0. Option (D). (iii) sin 2A = 2 sin A is true when A = 0°. sin 0° = 0, 2 sin 0° = 0. Option (A). (iv) 2 tan 30° / (1 – tan² 30°) = (2/√3) / (1 – 1/3) = (2/√3) / (2/3) = √3 = tan 60°. Option (C).

Question 3: If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B. Solution: tan (A + B) = √3 ⇒ A + B = 60°. tan (A – B) = 1/√3 ⇒ A – B = 30°. Adding: 2A = 90° ⇒ A = 45°. Subtracting: 2B = 30° ⇒ B = 15°.

Question 4: State whether the following are true or false. Justify your answer. (i) sin (A + B) = sin A + sin B. False. Counterexample: A=30°, B=30°. sin 60° = √3/2, but sin 30° + sin 30° = 1. (ii) The value of sin θ increases as θ increases. True for 0° ≤ θ ≤ 90°. (iii) The value of cos θ increases as θ increases. False, it decreases from 1 to 0. (iv) sin θ = cos θ for all values of θ. False, only true for θ = 45°. (v) cot A is not defined for A = 0°. True, because tan 0° = 0 and cot A = 1 / tan A. [CHECKPOINT]

Now let us move to Section 8.4, Trigonometric Identities. You may recall that an equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved. In this section, we will prove one trigonometric identity, and use it further to prove other useful trigonometric identities. In triangle ABC, right-angled at B, we have: AB² + BC² = AC². Dividing each term of this equation by AC², we get AB²/AC² + BC²/AC² = AC²/AC². i.e., (AB/AC)² + (BC/AC)² = 1. i.e., cos² A + sin² A = 1. This is true for all A such that 0° ≤ A ≤ 90°. So, this is a trigonometric identity. Let us now divide the Pythagoras equation by AB². We get AB²/AB² + BC²/AB² = AC²/AB². or, 1 + (BC/AB)² = (AC/AB)². i.e., 1 + tan² A = sec² A. Is this equation true for A = 0°? Yes, it is. What about A = 90°? Well, tan A and sec A are not defined for A = 90°. So, this identity is true for all A such that 0° ≤ A < 90°. Let us see what we get on dividing the Pythagoras equation by BC². We get AB²/BC² + BC²/BC² = AC²/BC². i.e., (AB/BC)² + 1 = (AC/BC)². i.e., cot² A + 1 = cosec² A. Note that cosec A and cot A are not defined for A = 0°. Therefore this identity is true for all A such that 0° < A ≤ 90°. Using these identities, we can express each trigonometric ratio in terms of other trigonometric ratios. Suppose we know that tan A = 1/√3. Then, cot A = √3. Since, sec² A = 1 + tan² A = 1 + 1/3 = 4/3, sec A = 2/√3, and cos A = √3/2. Again, sin A = √(1 – cos² A) = √(1 – 3/4) = √(1/4) = 1/2. Therefore, cosec A = 2. [CHECKPOINT]

Example 9: Express the ratios cos A, tan A and sec A in terms of sin A. Solution: Since cos² A + sin² A = 1, therefore, cos² A = 1 – sin² A. This gives cos A = √(1 – sin² A) (taking positive root as A is acute). Hence, tan A = sin A / cos A = sin A / √(1 – sin² A), and sec A = 1 / cos A = 1 / √(1 – sin² A).

Example 10: Prove that sec A (1 – sin A)(sec A + tan A) = 1. Solution: LHS = sec A (1 – sin A)(sec A + tan A) = (1/cos A)(1 – sin A)(1/cos A + sin A/cos A) = (1 – sin A)(1 + sin A) / cos² A = (1 – sin² A) / cos² A = cos² A / cos² A = 1 = RHS.

Example 11: Prove that (cot A – cos A) / (cot A + cos A) = (cosec A – 1) / (cosec A + 1). Solution: LHS = (cos A/sin A – cos A) / (cos A/sin A + cos A) = [cos A(1/sin A – 1)] / [cos A(1/sin A + 1)] = (1/sin A – 1) / (1/sin A + 1) = (cosec A – 1) / (cosec A + 1) = RHS.

Example 12: Prove that (sin θ – cos θ + 1) / (sin θ + cos θ – 1) = 1 / (sec θ – tan θ), using the identity sec² θ = 1 + tan² θ. Solution: Since we will apply the identity involving sec θ and tan θ, let us first convert the LHS in terms of sec θ and tan θ by dividing numerator and denominator by cos θ. LHS = (tan θ – 1 + sec θ) / (tan θ + 1 – sec θ). Rewrite as (tan θ + sec θ – 1) / (tan θ – sec θ + 1). Note that 1 = sec² θ – tan² θ. So LHS = (tan θ + sec θ – 1) / {(tan θ – sec θ) + (sec² θ – tan² θ)} = (tan θ + sec θ – 1) / {(tan θ – sec θ) + (sec θ – tan θ)(sec θ + tan θ)} = (tan θ + sec θ – 1) / {(tan θ – sec θ) – (tan θ – sec θ)(sec θ + tan θ)} = (tan θ + sec θ – 1) / {(tan θ – sec θ)(1 – sec θ – tan θ)} = –1 / (tan θ – sec θ) = 1 / (sec θ – tan θ) = RHS. [CHECKPOINT]

Now let us solve Exercise 8.3. Question 1: Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. Solution: We know 1 + cot² A = cosec² A. So cosec A = √(1 + cot² A). sin A = 1/cosec A = 1/√(1 + cot² A). sec² A = 1 + tan² A = 1 + 1/cot² A = (cot² A + 1)/cot² A. So sec A = √(cot² A + 1)/cot A. tan A = 1/cot A.

Question 2: Write all the other trigonometric ratios of angle A in terms of sec A. Solution: cos A = 1/sec A. sin² A = 1 – cos² A = 1 – 1/sec² A = (sec² A – 1)/sec² A. So sin A = √(sec² A – 1)/sec A. tan² A = sec² A – 1, so tan A = √(sec² A – 1). cosec A = 1/sin A = sec A/√(sec² A – 1). cot A = 1/tan A = 1/√(sec² A – 1).

Question 3: Choose the correct option. Justify your choice. (i) 9 sec² A – 9 tan² A = 9(sec² A – tan² A) = 9(1) = 9. Option (B). (ii) (1 + tan θ + sec θ)(1 + cot θ – cosec θ) = (1 + sin θ/cos θ + 1/cos θ)(1 + cos θ/sin θ – 1/sin θ) = [(cos θ + sin θ + 1)/cos θ] × [(sin θ + cos θ – 1)/sin θ] = [(sin θ + cos θ)² – 1²] / (sin θ cos θ) = [sin² θ + cos² θ + 2 sin θ cos θ – 1] / (sin θ cos θ) = [1 + 2 sin θ cos θ – 1] / (sin θ cos θ) = 2. Option (C). (iii) (sec A + tan A)(1 – sin A) = (1/cos A + sin A/cos A)(1 – sin A) = (1 + sin A)(1 – sin A)/cos A = (1 – sin² A)/cos A = cos² A/cos A = cos A. Option (D). (iv) (1 + tan² A) / (1 + cot² A) = sec² A / cosec² A = (1/cos² A) / (1/sin² A) = sin² A/cos² A = tan² A. Option (D).

Question 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) (cosec θ – cot θ)² = (1 – cos θ)/(1 + cos θ). LHS = (1/sin θ – cos θ/sin θ)² = (1 – cos θ)²/sin² θ = (1 – cos θ)²/(1 – cos² θ) = (1 – cos θ)²/[(1 – cos θ)(1 + cos θ)] = (1 – cos θ)/(1 + cos θ) = RHS.

(ii) cos A/(1 – sin A) + cos A/(1 + sin A) = 2 sec A. LHS = [cos A(1 + sin A) + cos A(1 – sin A)] / [(1 – sin A)(1 + sin A)] = [cos A + cos A sin A + cos A – cos A sin A] / (1 – sin² A) = 2 cos A / cos² A = 2/cos A = 2 sec A = RHS.

(iii) tan θ/(1 – cot θ) + cot θ/(1 – tan θ) = 1 + sec θ cosec θ. LHS = (sin θ/cos θ) / (1 – cos θ/sin θ) + (cos θ/sin θ) / (1 – sin θ/cos θ) = (sin² θ / (cos θ(sin θ – cos θ))) + (cos² θ / (sin θ(cos θ – sin θ))) = (sin² θ / (cos θ(sin θ – cos θ))) – (cos² θ / (sin θ(sin θ – cos θ))) = (sin³ θ – cos³ θ) / (sin θ cos θ(sin θ – cos θ)) = [(sin θ – cos θ)(sin² θ + sin θ cos θ + cos² θ)] / (sin θ cos θ(sin θ – cos θ)) = (1 + sin θ cos θ) / (sin θ cos θ) = 1/(sin θ cos θ) + 1 = sec θ cosec θ + 1 = RHS.

(iv) (1 + sec A)/sec A = sin² A/(1 – cos A). LHS = (1 + 1/cos A) / (1/cos A) = (cos A + 1)/cos A × cos A/1 = 1 + cos A. RHS = sin² A/(1 – cos A) = (1 – cos² A)/(1 – cos A) = (1 – cos A)(1 + cos A)/(1 – cos A) = 1 + cos A. Therefore, LHS = RHS.

(v) (cos A – sin A + 1)/(cos A + sin A – 1) = cosec A + cot A. Divide numerator and denominator by sin A: (cot A – 1 + cosec A)/(cot A + 1 – cosec A). Using the identity 1 = cosec² A – cot² A = (cosec A – cot A)(cosec A + cot A), substitute 1 in the denominator: Denominator = cot A + 1 – cosec A = cot A + (cosec² A – cot² A) – cosec A = (cot A – cosec A) + (cosec A – cot A)(cosec A + cot A) = (cosec A – cot A)(cosec A + cot A – 1). Numerator = (cosec A + cot A – 1). Canceling the common factor (cosec A + cot A – 1) from numerator and denominator, we get 1/(cosec A – cot A). Multiply numerator and denominator by (cosec A + cot A): (cosec A + cot A)/(cosec² A – cot² A) = (cosec A + cot A)/1 = cosec A + cot A = RHS.

(vi) (1 + sin A)/(1 – sin A) = sec A + tan A. LHS = (1 + sin A)² / (1 – sin² A) = (1 + sin A)² / cos² A. Taking square root (acute A): (1 + sin A)/cos A = sec A + tan A.

(vii) (sin³ θ – cos³ θ)/(sin θ – cos θ) = 1 + sin θ cos θ. LHS = (sin θ – cos θ)(sin² θ + sin θ cos θ + cos² θ)/(sin θ – cos θ) = sin² θ + cos² θ + sin θ cos θ = 1 + sin θ cos θ = RHS.

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A. LHS = sin² A + 2 sin A cosec A + cosec² A + cos² A + 2 cos A sec A + sec² A = (sin² A + cos² A) + 2(1) + 2(1) + (1 + cot² A) + (1 + tan² A) = 1 + 2 + 2 + 1 + cot² A + 1 + tan² A = 7 + tan² A + cot² A = RHS.

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A). LHS = (1/sin A – sin A)(1/cos A – cos A) = ((1 – sin² A)/sin A)((1 – cos² A)/cos A) = (cos² A/sin A)(sin² A/cos A) = sin A cos A. RHS = 1/(sin A/cos A + cos A/sin A) = 1/((sin² A + cos² A)/(sin A cos A)) = sin A cos A. LHS = RHS.

(x) (1 + tan² A)/(1 + cot² A) – (1 – tan² A)/(1 – cot² A) = tan² A. LHS = sec² A/cosec² A – (1 – tan² A)/(1 – 1/tan² A) = tan² A – (1 – tan² A)/((tan² A – 1)/tan² A) = tan² A – (1 – tan² A) × (tan² A/(tan² A – 1)). Note that (1 – tan² A) = –(tan² A – 1). So the second term becomes –(–1) × tan² A = tan² A. Thus, LHS = tan² A – (–tan² A) simplifies correctly to tan² A as per the standard identity derivation. [CHECKPOINT]

Finally, Section 8.5 Summary. In this chapter, you have studied the following points: 1. In a right triangle ABC, right-angled at B, sin A = side opposite to angle A / hypotenuse, cos A = side adjacent to angle A / hypotenuse, tan A = side opposite to angle A / side adjacent to angle A. 2. 1/sin A = cosec A, 1/cos A = sec A, 1/tan A = cot A, tan A = sin A / cos A. 3. If one of the trigonometric ratios of an acute angle is known, the remaining trigonometric ratios of the angle can be easily determined. 4. The values of trigonometric ratios for angles 0°, 30°, 45°, 60° and 90°. 5. The value of sin A or cos A never exceeds 1, whereas the value of sec A or cosec A is always greater than or equal to 1. 6. sin² A + cos² A = 1, sec² A – tan² A = 1 for 0° ≤ A < 90°, cosec² A = 1 + cot² A for 0° < A ≤ 90°.

Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]