Welcome to the ultimate guide for Class 10 Mathematics Chapter 8: Introduction to Trigonometry. Trigonometry is a foundational branch of mathematics that studies the relationships between the side lengths and angles of right-angled triangles. Below you will find important formulas followed by step-by-step solutions for all exercises in Chapter 8.

Exercise 8.1 Solutions

Q1. In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C

Solution:
In right Δ ABC, by Pythagoras theorem:
AC² = AB² + BC² = (24)² + (7)² = 576 + 49 = 625
AC = √625 = 25 cm.
(i) For angle A: Opposite side = BC = 7 cm, Adjacent side = AB = 24 cm, Hypotenuse = AC = 25 cm.
sin A = BC/AC = 7/25
cos A = AB/AC = 24/25
(ii) For angle C: Opposite side = AB = 24 cm, Adjacent side = BC = 7 cm, Hypotenuse = AC = 25 cm.
sin C = AB/AC = 24/25
cos C = BC/AC = 7/25

Q2. In Fig. 8.13, find tan P – cot R. (Given: PQ = 12 cm, PR = 13 cm)

Solution:
In right Δ PQR, right-angled at Q:
PR² = PQ² + QR² ⇒ 13² = 12² + QR² ⇒ 169 = 144 + QR² ⇒ QR² = 25 ⇒ QR = 5 cm.
tan P = Opposite/Adjacent = QR/PQ = 5/12
cot R = Adjacent/Opposite = QR/PQ = 5/12
Therefore, tan P – cot R = 5/12 – 5/12 = 0.

Q3. If sin A = 3/4, calculate cos A and tan A.

Solution:
sin A = Opposite/Hypotenuse = 3/4.
Let Opposite side = 3k and Hypotenuse = 4k.
By Pythagoras theorem: Adjacent side = √[(4k)² – (3k)²] = √(16k² – 9k²) = k√7.
cos A = Adjacent/Hypotenuse = (k√7)/(4k) = √7/4
tan A = Opposite/Adjacent = (3k)/(k√7) = 3/√7

Q4. Given 15 cot A = 8, find sin A and sec A.

Solution:
15 cot A = 8 ⇒ cot A = 8/15 = Adjacent/Opposite.
Let Adjacent = 8k, Opposite = 15k.
Hypotenuse = √[(8k)² + (15k)²] = √(64k² + 225k²) = √289k² = 17k.
sin A = Opposite/Hypotenuse = 15k/17k = 15/17
sec A = Hypotenuse/Adjacent = 17k/8k = 17/8

Q5. Given sec θ = 13/12, calculate all other trigonometric ratios.

Solution:
sec θ = Hypotenuse/Adjacent = 13/12.
Let Hypotenuse = 13k, Adjacent = 12k.
Opposite = √[(13k)² – (12k)²] = √(169k² – 144k²) = √25k² = 5k.
1. sin θ = Opposite/Hypotenuse = 5/13
2. cos θ = Adjacent/Hypotenuse = 12/13
3. tan θ = Opposite/Adjacent = 5/12
4. cosec θ = Hypotenuse/Opposite = 13/5
5. cot θ = Adjacent/Opposite = 12/5

Q6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Solution:
Consider a right-angled triangle ABC, right-angled at C.
cos A = AC/AB, cos B = BC/AB.
Given cos A = cos B ⇒ AC/AB = BC/AB ⇒ AC = BC.
Since sides opposite to equal angles are equal in a triangle (or angles opposite to equal sides are equal), AC = BC implies ∠A = ∠B. Proved.

Q7. If cot θ = 7/8, evaluate: (i) [(1 + sin θ)(1 – sin θ)] / [(1 + cos θ)(1 – cos θ)] (ii) cot² θ

Solution:
(i) [(1 + sin θ)(1 – sin θ)] / [(1 + cos θ)(1 – cos θ)] = (1 – sin² θ) / (1 – cos² θ) = cos² θ / sin² θ = cot² θ.
Since cot θ = 7/8, cot² θ = (7/8)² = 49/64.
(ii) cot² θ = (7/8)² = 49/64.

Q8. If 3 cot A = 4, check whether (1 – tan² A)/(1 + tan² A) = cos² A – sin² A or not.

Solution:
3 cot A = 4 ⇒ cot A = 4/3 ⇒ tan A = 3/4.
Adjacent = 4k, Opposite = 3k ⇒ Hypotenuse = √(16k² + 9k²) = 5k.
sin A = 3/5, cos A = 4/5.
LHS = (1 – (3/4)²) / (1 + (3/4)²) = (1 – 9/16) / (1 + 9/16) = (7/16) / (25/16) = 7/25.
RHS = cos² A – sin² A = (4/5)² – (3/5)² = 16/25 – 9/25 = 7/25.
Since LHS = RHS, Yes, it is equal.

Q9. 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 ⇒ Opposite (BC) = 1k, Adjacent (AB) = √3k.
AC = √[(1k)² + (√3k)²] = √(1k² + 3k²) = √4k² = 2k.
sin A = 1/2, cos A = √3/2.
For angle C: Opposite = √3k, Adjacent = 1k, Hypotenuse = 2k.
sin C = √3/2, cos C = 1/2.
(i) sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 4/4 = 1.
(ii) cos A cos C – sin A sin C = (√3/2)(1/2) – (1/2)(√3/2) = √3/4 – √3/4 = 0.

Q10. In Δ 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 cm, then PR = (25 – x) cm.
By Pythagoras theorem in Δ PQR:
PR² = PQ² + QR² ⇒ (25 – x)² = 5² + x²
⇒ 625 – 50x + x² = 25 + x² ⇒ 50x = 600 ⇒ x = 12.
So, QR = 12 cm, PR = 25 – 12 = 13 cm, PQ = 5 cm.
sin P = QR/PR = 12/13
cos P = PQ/PR = 5/13
tan P = QR/PQ = 12/5

Q11. State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1. — False (tan 60° = √3 ≈ 1.732 > 1).
(ii) sec A = 12/5 for some value of angle A. — True (sec A ≥ 1 always, and 12/5 = 2.4 > 1).
(iii) cos A is the abbreviation used for the cosecant of angle A. — False (cos A is short for cosine; cosecant is cosec A).
(iv) cot A is the product of cot and A. — False (cot A is the cotangent ratio of angle A, not multiplication).
(v) sin θ = 4/3 for some angle θ. — False (sin θ cannot be greater than 1, but 4/3 = 1.33 > 1).

Exercise 8.2 Solutions

Q1. 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) = (1/√2) / [(2 + 2√3)/√3] = √3 / [√2(2 + 2√3)] = √3 / (2√2 + 2√6).
Rationalizing denominator: [√3(2√6 – 2√2)] / [(2√6 + 2√2)(2√6 – 2√2)] = (2√18 – 2√6) / (24 – 8) = (6√2 – 2√6) / 16 = (3√2 – √6) / 8.

(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) = [(3√3 – 4)/2√3] / [(3√3 + 4)/2√3] = (3√3 – 4) / (3√3 + 4).
Rationalizing: (3√3 – 4)² / (27 – 16) = (27 + 16 – 24√3) / 11 = (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 + 64 – 12)/12] / 1 = 67/12.

Q2. Choose the correct option and justify your choice:

(i) 2 tan 30° / (1 + tan² 30°) =
= [2(1/√3)] / [1 + (1/√3)²] = (2/√3) / (4/3) = 6 / 4√3 = √3/2 = (A) sin 60°.

(ii) (1 – tan² 45°) / (1 + tan² 45°) =
= (1 – 1²) / (1 + 1²) = 0 / 2 = 0 = (D) 0.

(iii) sin 2A = 2 sin A is true when A =
When A = 0°: sin 0° = 0 and 2 sin 0° = 0. So (A) 0°.

(iv) 2 tan 30° / (1 – tan² 30°) =
= [2(1/√3)] / [1 – (1/√3)²] = (2/√3) / (2/3) = √3 = (C) tan 60°.

Q3. 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° --- (Equation 1)
tan (A – B) = 1/√3 ⇒ A – B = 30° --- (Equation 2)
Adding (1) and (2): 2A = 90° ⇒ A = 45°.
Substitute A in (1): 45° + B = 60° ⇒ B = 15°.

Q4. State whether true or false:

(i) sin (A + B) = sin A + sin B — False (e.g., A=30°, B=60°: sin 90°=1, but sin 30°+sin 60°=1/2+√3/2 ≠ 1).
(ii) The value of sin θ increases as θ increases — True (goes from 0 to 1 as θ goes from 0° to 90°).
(iii) The value of cos θ increases as θ increases — False (decreases from 1 to 0).
(iv) sin θ = cos θ for all values of θ — False (true only for θ = 45°).
(v) cot A is not defined for A = 0° — True (cot 0° = cos 0°/sin 0° = 1/0, undefined).

Exercise 8.3 Solutions

Q1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution:
1. cosec² A = 1 + cot² A ⇒ cosec A = √(1 + cot² A). Since sin A = 1/cosec A, sin A = 1 / √(1 + cot² A).
2. tan A = 1 / cot A.
3. sec² A = 1 + tan² A = 1 + 1/cot² A = (cot² A + 1)/cot² A ⇒ sec A = √(cot² A + 1) / cot A.

Q2. Write all other trigonometric ratios of ∠A in terms of sec A.

Solution:
1. cos A = 1 / sec A
2. sin² A = 1 – cos² A = 1 – 1/sec² A = (sec² A – 1)/sec² A ⇒ sin A = √(sec² A – 1) / sec A
3. tan A = √(sec² A – 1) (since 1 + tan² A = sec² A)
4. cosec A = 1 / sin A = sec A / √(sec² A – 1)
5. cot A = 1 / tan A = 1 / √(sec² A – 1)

Q3. Choose the correct option:

(i) 9 sec² A – 9 tan² A =
= 9(sec² A – tan² A) = 9(1) = (B) 9.

(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) = (1 + 2 sin θ cos θ – 1) / (sin θ cos θ) = 2 sin θ cos θ / sin θ cos θ = (C) 2.

(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 = (D) cos A.

(iv) (1 + tan² A) / (1 + cot² A) =
= sec² A / cosec² A = (1/cos² A) / (1/sin² A) = sin² A / cos² A = (D) tan² A.

Q4. Prove the following identities:

(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. Proved.

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

(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 + cos² θ + sin θ cos 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. Proved.

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

(v) (cos A – sin A + 1)/(cos A + sin A – 1) = cosec A + cot A
Divide numerator & denominator by sin A:
LHS = (cot A – 1 + cosec A)/(cot A + 1 – cosec A) = [(cot A + cosec A) – (cosec² A – cot² A)] / (cot A + 1 – cosec A)
= [(cosec A + cot A)(1 – cosec A + cot A)] / (1 – cosec A + cot A) = cosec A + cot A = RHS. Proved.

(vi) √[(1 + sin A)/(1 – sin A)] = sec A + tan A
LHS = √[((1 + sin A)(1 + sin A)) / ((1 – sin A)(1 + sin A))] = √[(1 + sin A)² / (1 – sin² A)] = √[(1 + sin A)² / cos² A] = (1 + sin A)/cos A = 1/cos A + sin A/cos A = sec A + tan A = RHS. Proved.

(vii) (sin θ – 2 sin³ θ) / (2 cos³ θ – cos θ) = tan θ
LHS = [sin θ(1 – 2 sin² θ)] / [cos θ(2 cos² θ – 1)] = tan θ * [1 – 2(1 – cos² θ)] / [2 cos² θ – 1] = tan θ * [2 cos² θ – 1] / [2 cos² θ – 1] = tan θ = RHS. Proved.

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
LHS = sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A
= (sin² A + cos² A) + 2(1) + 2(1) + cosec² A + sec² A
= 1 + 4 + (1 + cot² A) + (1 + tan² A) = 7 + tan² A + cot² A = RHS. Proved.

(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. Proved.

(x) (1 + tan² A)/(1 + cot² A) = [(1 – tan A)/(1 – cot A)]² = tan² A
Part 1: (1 + tan² A)/(1 + cot² A) = sec² A / cosec² A = sin² A / cos² A = tan² A.
Part 2: [(1 – tan A)/(1 – 1/tan A)]² = [(1 – tan A) / ((tan A – 1)/tan A)]² = [–tan A]² = tan² A.
All parts equal tan² A. Proved.

🔍 Search more MP Board Solutions