Chapter 8: Introduction to Trigonometry (NCERT Solutions)

Exercise 8.1: Trigonometric Ratios

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

AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625.
AC = √625 = 25 cm.

(i) sin A = BC/AC = 7/25. cos A = AB/AC = 24/25.

(ii) sin C = AB/AC = 24/25. cos C = BC/AC = 7/25.

Q2. In Fig. 8.13, find tan P - cot R.

PQ = 12 cm, PR = 13 cm.
QR = √(13² - 12²) = √(169 - 144) = √25 = 5 cm.
tan P = QR/PQ = 5/12.
cot R = QR/PQ = 5/12.
tan P - cot R = 5/12 - 5/12 = 0.

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

Let BC = 3k, AC = 4k.
AB = √((4k)² - (3k)²) = √(16k² - 9k²) = √7k.
cos A = AB/AC = √7k/4k = √7/4.
tan A = BC/AB = 3k/√7k = 3/√7.

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

cot A = 8/15. AB=8k, BC=15k.
AC = √(64k² + 225k²) = √289k² = 17k.
sin A = BC/AC = 15/17.
sec A = AC/AB = 17/8.

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

Hyp = 13k, Adj = 12k. Opp = √(169-144) = 5k.
sin θ = 5/13, cos θ = 12/13.
tan θ = 5/12, cot θ = 12/5.
cosec θ = 13/5.

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

In ΔABC (assuming right angle at C for simplicity, or general method).
cos A = AC/AB. cos B = BC/AB.
AC/AB = BC/AB ⇒ AC = BC.
Angles opposite to equal sides are equal. ∴ ∠A = ∠B.

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

(i) (1 - sin² θ) / (1 - cos² θ) = cos² θ / sin² θ = 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.

cot A = 4/3 ⇒ tan A = 3/4.
LHS = (1 - 9/16) / (1 + 9/16) = (7/16) / (25/16) = 7/25.
sin A = 3/5, cos A = 4/5.
RHS = (4/5)² - (3/5)² = 16/25 - 9/25 = 7/25.
LHS = RHS. Yes.

Q9. In ΔABC, 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

tan A = 1/√3 ⇒ A = 30°. So C = 60°.
(i) sin 30 cos 60 + cos 30 sin 60 = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.
(ii) cos 30 cos 60 - sin 30 sin 60 = (√3/2)(1/2) - (1/2)(√3/2) = 0.

Q10. In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine sin P, cos P and tan P.

Let QR = x, PR = 25-x.
(25-x)² = 5² + x².
625 + x² - 50x = 25 + x².
50x = 600 ⇒ x = 12.
QR = 12, PR = 13.
sin P = 12/13, cos P = 5/13, tan P = 12/5.

Q11. State whether the following are true or false.

(i) Value of tan A is always less than 1. False (e.g. tan 60 = √3 ≈ 1.732).

(ii) sec A = 12/5 for some value of angle A. True (sec A ≥ 1).

(iii) cos A is abbreviation for cosecant. False (cosine).

(iv) cot A is product of cot and A. False.

(v) sin θ = 4/3 for some angle θ. False (sin θ ≤ 1).

Exercise 8.2: Trigonometric Ratios of Specific Angles

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)(1+√3)] = √3 / [2√2(√3+1)].
Rationalise: √3(√3-1) / [2√2(2)] = (3-√3) / 4√2
Rationalise √2: √2(3-√3) / 8 = (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) / (3√3 + 4).
Rationalise: (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)
Num: 5(1/2)² + 4(2/√3)² - 1 = 5/4 + 16/3 - 1 = (15 + 64 - 12)/12 = 67/12.
Den: 1.
67/12.

Q2. Choose the correct option.

(i) 2 tan 30 / (1 + tan² 30): 2(1/√3) / (1 + 1/3) = (2/√3) / (4/3) = 3 / 2√3 = √3/2 = sin 60 (A).

(ii) (1 - tan² 45) / (1 + tan² 45): 0/2 = 0 (D).

(iii) sin 2A = 2 sin A: Only true for A = 0° (A).

(iv) 2 tan 30 / (1 - tan² 30): 2(1/√3) / (2/3) = √3 = tan 60 (C).

Q3. If tan(A+B) = √3 and tan(A-B) = 1/√3... find A and B.

A+B = 60°.
A-B = 30°.
Add: 2A = 90 ⇒ A = 45°.
B = 15°.

Q4. True or False.

(i) sin(A+B) = sinA + sinB. False.

(ii) Value of sin θ increases as θ increases. True (0 to 90).

(iii) Value of cos θ increases as θ increases. False (decreases).

(iv) sin θ = cos θ for all θ. False (only 45).

(v) cot A is not defined for A=0. True.

Exercise 8.3: Complementary Angles

Q1. Evaluate:

(i) sin 18 / cos 72: sin 18 / sin(90-72) = sin 18 / sin 18 = 1.

(ii) tan 26 / cot 64: tan 26 / tan 26 = 1.

(iii) cos 48 - sin 42: sin 42 - sin 42 = 0.

(iv) cosec 31 - sec 59: sec 59 - sec 59 = 0.

Q2. Show that:

(i) tan 48 tan 23 tan 42 tan 67 = 1
(tan 48 tan 42) (tan 23 tan 67) = (tan 48 cot 48) (tan 23 cot 23) = 1 × 1 = 1.

(ii) cos 38 cos 52 - sin 38 sin 52 = 0
cos 38 sin 38 - sin 38 cos 38 = 0.

Q3. If tan 2A = cot(A-18), find A.

cot(90-2A) = cot(A-18).
90 - 2A = A - 18 ⇒ 3A = 108 ⇒ A = 36°.

Q4. If tan A = cot B, prove A + B = 90.

tan A = tan(90-B) ⇒ A = 90 - B ⇒ A + B = 90°.

Q5. If sec 4A = cosec(A-20), find A.

cosec(90-4A) = cosec(A-20).
90 - 4A = A - 20 ⇒ 5A = 110 ⇒ A = 22°.

Q6. If A, B, C are interior angles, show sin((B+C)/2) = cos A/2.

B+C = 180-A.
sin((180-A)/2) = sin(90 - A/2) = cos A/2.

Q7. Express sin 67 + cos 75 in terms of angles between 0 and 45.

cos(90-67) + sin(90-75) = cos 23 + sin 15.

Exercise 8.4: Trigonometric Identities

Q1. Express sin A, sec A, tan A in terms of cot A.

sin A: 1/√(1+cot² A).
sec A: √(1+tan² A) = √(1 + 1/cot² A) = √(cot² A + 1) / cot A.
tan A: 1/cot A.

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

cos A = 1/sec A.
sin A = √(1-cos² A) = √(1-1/sec² A) = √(sec² A-1) / sec A.
tan A = √(sec² A - 1).
cosec A = sec A / √(sec² A - 1).
cot A = 1 / √(sec² A - 1).

Q3. Evaluate.

(i) (sin² 63 + sin² 27) / (cos² 17 + cos² 73)
Num: sin² 63 + cos² 63 = 1.
Den: sin² 73 + cos² 73 = 1.
Result: 1.

(ii) sin 25 cos 65 + cos 25 sin 65
sin 25 sin 25 + cos 25 cos 25 = sin² 25 + cos² 25 = 1.

Q4. Choose correct option.

(i) 9 sec² A - 9 tan² A: 9(1) = 9 (B).

(ii) (1 + tan + sec)(1 + cot - cosec):
(1 + s/c + 1/c)(1 + c/s - 1/s) = [(c+s+1)/c][(s+c-1)/s] = [(c+s)² - 1]/sc = (c²+s²+2sc-1)/sc = 2sc/sc = 2 (C).

(iii) (sec A + tan A)(1 - sin A): (1/c + s/c)(1-s) = (1+s)(1-s)/c = (1-s²)/c = c²/c = cos A = (D).

(iv) (1+tan² A)/(1+cot² A): sec² A / cosec² A = (1/c²) / (1/s²) = s²/c² = tan² A = (D).

Q5. Prove the following identities.

(i) (cosec - cot)² = (1-cos)/(1+cos)
LHS = (1/s - c/s)² = (1-c)²/s² = (1-c)²/(1-c²) = (1-c)/(1+c). Proved.

(ii) cos/(1+sin) + (1+sin)/cos = 2 sec A
LHS = (c² + (1+s)²) / c(1+s) = (c² + 1 + s² + 2s)/c(1+s) = (2 + 2s)/c(1+s) = 2(1+s)/c(1+s) = 2/c = 2 sec A. Proved.

(iii) tan/(1-cot) + cot/(1-tan) = 1 + sec cosec
Convert to sin/cos. (s/c)/(1-c/s) + (c/s)/(1-s/c).
= s²/c(s-c) + c²/s(c-s) = s²/c(s-c) - c²/s(s-c)
= (s³-c³)/sc(s-c) = (s-c)(s²+c²+sc)/sc(s-c) = (1+sc)/sc = 1/sc + 1 = 1 + sec cosec. Proved.

(iv) (1+sec)/sec = sin²/(1-cos)
LHS = (1+1/c)/(1/c) = c+1.
RHS = (1-c²)/(1-c) = 1+c. LHS=RHS.

(v) (cos-sin+1)/(cos+sin-1) = cosec + cot
Divide num/den by sin.
(cot-1+cosec)/(cot+1-cosec) = (cot+cosec - (cosec²-cot²))/(cot-cosec+1)
= (cot+cosec)(1 - (cosec-cot)) / (cot-cosec+1) = cot + cosec. Proved.

(vi) √((1+sin)/(1-sin)) = sec + tan
Multiply by √(1+sin).
√(1+sin)²/√(1-sin²) = (1+sin)/cos = sec + tan. Proved.

(vii) (sin - 2sin³)/(2cos³ - cos) = tan
s(1-2s²) / c(2c²-1).
1-2s² = 1 - 2(1-c²) = 2c²-1.
s/c = tan. Proved.

(viii) (sin+cosec)² + (cos+sec)² = 7 + tan² + cot²
s²+c² + 2sc s + c²+s² + 2cs.
1 + (1+cot²) + 2(1) + (1+tan²) + 2(1) = 1+1+2+1+2 + tan²+cot² = 7 + tan² + cot². Proved.

(ix) (cosec-sin)(sec-cos) = 1/(tan+cot)
LHS = (1/s - s)(1/c - c) = (c²/s)(s²/c) = sc.
RHS = 1/(s/c + c/s) = 1 / ((s²+c²)/sc) = sc. Proved.

(x) (1+tan²)/(1+cot²) = ( (1-tan)/(1-cot) )² = tan²
LHS = sec²/cosec² = tan².
Mid = ((1-t)/(1-1/t))² = ((1-t)/((t-1)/t))² = (-t)² = tan². Proved.

Introduction to Trigonometry - RD Sharma Important Questions

Q1. In ΔABC, right angled at B, if tan A = 1/√3, find the value of sin A cos C + cos A sin C.

tan A = 1/√3 ⇒ A = 30°.
In right triangle, A+C=90, so C=60°.
Expression = sin(A+C) = sin 90° = 1.

Q2. If √3 tan θ = 1, then find the value of sin² θ - cos² θ.

tan θ = 1/√3 ⇒ θ = 30°.
sin² 30 - cos² 30 = (1/2)² - (√3/2)² = 1/4 - 3/4 = -2/4 = -1/2.

Q3. If 7 sin² θ + 3 cos² θ = 4, show that tan θ = 1/√3.

4 sin² θ + 3(sin² θ + cos² θ) = 4.
4 sin² θ + 3 = 4.
4 sin² θ = 1 ⇒ sin θ = 1/2.
θ = 30° ⇒ tan 30 = 1/√3.

Q4. Prove that: √((1+cos θ)/(1-cos θ)) = cosec θ + cot θ.

Multiply num and den by √(1+cos θ).
= (1+cos θ) / √(1-cos² θ)
= (1+cos θ) / sin θ
= 1/sin θ + cos θ/sin θ = cosec θ + cot θ.

Q5. If x = a cos θ - b sin θ and y = a sin θ + b cos θ, prove that x² + y² = a² + b².

x² = a²c² + b²s² - 2absc.
y² = a²s² + b²c² + 2absc.
Add: a²(c²+s²) + b²(s²+c²) = a² + b².

Q6. Evaluate: 4(sin⁴ 30 + cos⁴ 60) - 3(cos² 45 - sin² 90).

4((1/2)⁴ + (1/2)⁴) - 3((1/√2)² - 1²).
4(1/16 + 1/16) - 3(1/2 - 1).
4(2/16) - 3(-1/2) = 1/2 + 3/2 = 2.

Q7. If cos θ + sin θ = √2 cos θ, show that cos θ - sin θ = √2 sin θ.

Square given: c² + s² + 2sc = 2c².
s² + 2sc = c² ⇒ s² = c² - 2sc.
Consider (c-s)² = c² + s² - 2sc = (s²+2sc) + s² - 2sc = 2s².
c-s = √2 sin θ.

Q8. If sec θ + tan θ = p, show that (p²-1)/(p²+1) = sin θ.

p = (1+s)/c.
p² = (1+s)²/c².
Simplify expression using componendo-dividendo rules or direct substitution.
Result is sin θ. Verified.

Q9. Prove that: tan² A - tan² B = (sin² A - sin² B) / (cos² A cos² B).

LHS = s_A²/c_A² - s_B²/c_B².
= (s_A²c_B² - s_B²c_A²) / (c_A²c_B²).
Num = s_A²(1-s_B²) - s_B²(1-s_A²) = s_A² - s_B².
Hence proved.

Q10. Determine the value of x such that 2 cosec² 30 + x sin² 60 - 3/4 tan² 30 = 10.

2(2)² + x(√3/2)² - 3/4(1/√3)² = 10.
8 + 3x/4 - 3/4(1/3) = 10.
8 + 3x/4 - 1/4 = 10.
3x/4 = 2 + 1/4 = 9/4.
3x = 9 ⇒ x = 3.

Q11. Prove that: (1 + cot A - cosec A)(1 + tan A + sec A) = 2.

See NCERT Ex 8.4 Q4 (ii). Standard identity.
= 2.

Q12. If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1.

Square it: 1 + 2sc = 3 ⇒ 2sc = 2 ⇒ sc = 1.
LHS = s/c + c/s = (s²+c²)/sc = 1/sc = 1/1 = 1.

Q13. If A, B, C are interior angles of a triangle ABC, prove that tan((C+A)/2) = cot(B/2).

A+C = 180-B.
tan((180-B)/2) = tan(90 - B/2) = cot(B/2).

Q14. Find the value of sin² 5 + sin² 10 + ... + sin² 85 + sin² 90.

Series is 5, 10... 90 (18 terms).
sin² 90 = 1.
Pairs: (5, 85), (10, 80)... (40, 50). 8 pairs.
sin² x + sin² (90-x) = 1.
Sum = 8(1) + sin² 45 + 1.
= 9 + 0.5 = 9.5.

Q15. If tan A = n tan B and sin A = m sin B, prove that cos² A = (m²-1)/(n²-1).

cot B = n/tan A; cosec B = m/sin A.
cosec² B - cot² B = 1.
Substitute and solve for cos A. Result follows.

Q16. Prove: (sin A + sec A)² + (cos A + cosec A)² = (1 + sec A cosec A)².

Expand and simplify both sides.
LHS = s² + sec² + 2s sec + c² + cosec² + 2c cosec.
= 1 + sec² + cosec² + 2 tan + 2 cot.
RHS = 1 + sec² cosec² + 2 sec cosec.
Both reduce to same expression.

Q17. Prove: (tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A.

Replace 1 with sec² - tan².
Num = (t+s) - (s-t)(s+t) = (t+s)(1-s+t).
Den = t-s+1.
Cancels out leaving t+s = s/c + 1/c = (1+s)/c. Proved.

Q18. If a cos θ + b sin θ = m and a sin θ - b cos θ = n, prove a² + b² = m² + n².

Square and add.
m² + n² = a²(c²+s²) + b²(s²+c²).
= a² + b².

Q19. Evaluate without tables: (cos 70 / sin 20) + (cos 55 cosec 35) / (tan 5 tan 25 tan 45 tan 65 tan 85).

1 + 1 / (1) = 2.

Q20. If sin θ + 2 cos θ = 1, prove that 2 sin θ - cos θ = 2.

Let 2s - c = x.
(s+2c)² + (2s-c)² = 1² + x².
s²+4c²+4sc + 4s²+c²-4sc = 1+x².
5(1) = 1 + x² ⇒ x² = 4 ⇒ x = 2 (taking positive).

Introduction to Trigonometry - Formulas & PYQs

Key Formulas

1. Trigonometric Ratios (Right Triangle)

sin A = Opp/Hyp

cos A = Adj/Hyp

tan A = Opp/Adj = sin A / cos A

cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A

2. Values Table

θ	0°	30°	45°	60°	90°
sin	0	1/2	1/√2	√3/2	1
cos	1	√3/2	1/√2	1/2	0
tan	0	1/√3	1	√3	ND

3. Trigonometric Identities

sin² A + cos² A = 1

1 + tan² A = sec² A

1 + cot² A = cosec² A

4. Complementary Angles

sin(90-A) = cos A

tan(90-A) = cot A

sec(90-A) = cosec A

Previous Year Questions (CBSE/JKBOSE)

Q1. If sin θ + sin² θ = 1, prove that cos² θ + cos⁴ θ = 1. (CBSE 2020)

sin θ = 1 - sin² θ = cos² θ.
LHS = cos² θ + (cos² θ)² = sin θ + sin² θ = 1.
Proved.

Q2. Evaluate: 2 tan² 45 + cos² 30 - sin² 60. (CBSE 2019)

2(1)² + (√3/2)² - (√3/2)² = 2 + 3/4 - 3/4 = 2.

Q3. If tan θ = 4/3, find (sin θ + cos θ)/(sin θ - cos θ). (CBSE 2018)

Divide by cos θ.
(t+1)/(t-1) = (4/3 + 1)/(4/3 - 1) = (7/3)/(1/3) = 7.

Q4. Prove (sin A - 2 sin³ A)/(2 cos³ A - cos A) = tan A. (CBSE 2018)

Standard textbook question. Factor out sin A and cos A. Cancel bracket terms.

📱 Practice MCQs for this topic inside our App