Maths Class 10th Introduction to Trigonometry

NCERT CBSE for Class 10 Maths Chapter 8 Introduction to Trigonometry Important Question

Introduction to Trigonometry Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
If tan θ + cot θ = 5, find the value of tan²θ + cotθ.
Year of Question:(2012)

Solution:
tan θ + cot θ = 5 . [Given
tan²θ + cot²θ + 2 tan θ cot θ = 25 . [Squaring both sides
tan²θ + cot²θ + 2 = 25
∴ tan²θ + cot2θ = 23

Question 2.
If sec 2A = cosec (A - 27°) where 2A is an acute angle, find the measure of ∠A.
Year of Question:(2012, 2017D)

Solution:
sec 2A = cosec (A - 27°)
cosec(90° - 2A) = cosec(A - 27°) .[∵ sec θ = cosec (90° - θ)
90° - 2A = A - 27°
90° + 27° = 2A + A
⇒3A = 117°
∴∠A = 117°/3 = 39°

Question 3.
If tan α = √3 and tan β = 1/√3,0 < α, β < 90°, find the value of cot (α + β).
Year of Question:(2012)

Solution:
tan α = √3 = tan 60° .(i)
tan β = 1/√3 = tan 30° .(ii)
Solving (i) & (ii), α = 60° and β = 30°
∴ cot (α + β) = cot (60° + 30°) = cot 90° = 0

Question 4.
If sin θ - cos θ = 0, find the value of sin4 θ + cos4 θ.
Year of Question:(2012, 2017D)

Solution:
sin θ - cos θ = 0 = sin θ = cos θ
⇒ sinθ/cosθ = 1 ⇒ tan θ = 1 ⇒ θ = 45°
Now, sin⁴θ + cos⁴θ
= sin⁴ 45° + cos⁴ 45°
= (1/√2)4+(1/√2)4=1/4+1/4=2/4=1/2

Question 5.
If sec θ + tan θ = 7, then evaluate sec θ - tan θ.
Year of Question:(2017OD)

Solution:
We know that,
sec²θ - tan²θ = 1
(sec θ + tanθ) (sec θ - tan θ) = 1
(7) (sec θ - tan θ) = 1 .[sec θ + tan θ = 7; (Given)
∴ sec θ - tan θ = 1/7

Question 6.
Evaluate: 10. 1-cot²45°/1+sin⁹⁰°.
Year of Question:(2014)

Solution:

Question 7.
If cosec θ = 5/4, find the value of cot θ.
Year of Question:(2014)

Solution:
We know that, cot²θ = cosec²θ - 1
= (5/4)² - 1 ⇒ 25/16 - 1 ⇒ 25-16/16
cot²θ = 9/16 i cot - = 34

Question 8.
If θ = 45°, then what is the value of 2 sec²θ + 3 cosec2θ -
Year of Question:(2014)

Solution:
2 sec²θ + 3 cosec²θ = 2 sec² 45° + 3 cosec² 45°
= 2(√2)² + 3 (√2)² = 4 + 6 = 10

Question 9.
If √3 sin θ = cos θ, find the value of 3cos²θ+2cosθ/3cosθ+2.
Year of Question:(2015)

Solution:
√3 sin θ = cos θ . [Given

Question 10.
Evaluate: sin² 19° + sin771°.
Year of Question:(2015)

Solution:
sin² 19° + sin² 71°
= sin²19° + sin² (90° - 19°).[∵ sin(90° - θ) = cos θ
= sin² 19° + cos² 19° = 1 .[∵ sin2 θ + cos² θ = 1

Question 11.
What happens to value of cos when increases from 0° to 90°?
Year of Question:(2015)

Solution:
cos 0° = 1, cos 90° = 0
When θ increases from 0° to 90°, the value of cos θ decreases from 1 to 0.

Question 12.
If tan θ= a/x, find the value of x/√a²+x².
Year of Question:(2013)

Solution:

Question 13.
If in a right angled ∆ABC, tan B = 12/5, then find sin B.
Year of Question:(2014)

Solution:
1st method:
tan B = 12/5 ∴ cot B = 5/12
cosec² B = 1 + cot² B
= 1 + [(5/12)²/latex]=1+[latex]
= 144+25/144=169/144
cosec B = 13/12 ∴ sin B = 12/13
2nd method:

tan B = 12/5
tan B = AC/BC
Let AC = 12k, BC = 5k
In rt. ∆ACB,
AB² = AC² + BC² .[Pythagoras theorem
AB² = (12k)² + (5k)²
AB² = 144k² + 25k²2 = 169k²
AB = 13k
∴sin B = AC/AB=12k/13k=12/13

Question 14.
If ∆ABC is right angled at B, what is the value of sin (A + C).
Year of Question:(2015)

Solution:

∠B = 90° .[Given
∠A + ∠B + ∠C = 180° .[Angle sum property of a ∆
∠A + ∠C + 90° = 180°
∠A + ∠C = 90°
∴ sin (A + C) = sin 90° = 1 .(taking sin both side

Introduction to Trigonometry Class 10 Important Questions Short Answer-I (2 Marks)

Question 15.
Evaluate: tan 15° . tan 25° , tan 60° . tan 65° . tan 75° - tan 30°.
Year of Question:(2013)

Solution:
tan 15°. tan 25°, tan 60°. tan 65°. tan 75° - tan 30°
= tan(90° - 75°) tan(90° - 65°). √3 . tan 65°. tan 75° - 1/√3

Question 16.
Express cot 75° + cosec 75° in terms of trigonometric ratios of angles between 0° and 30°.
Year of Question:(2013)

Solution:
cot 75° + cosec 75°
= cot(90° - 15°) + cosec(90° - 15°)
= tan 15° + sec 15° .[cot(90°-A) = tan A
cosec(90° - A) = sec A

Question 17.
If cos (A + B) = 0 and sin (A - B) = 3, then find the value of A and B where A and B are acute angles.
Year of Question:(2012)

Solution:

Putting the value of B in (i), we get
⇒ A = 30° + 30° = 60°
∴ A = 60°, B = 30°

Question 18.
If A, B and C are the interior angles of a ∆ABC, show that sin (A+B/2) = cos(c/2).
Year of Question:(2012)

Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180° .(Angle sum property of ∆
∠A + ∠B = 180° - ∠C

Question 19.
If x = p sec θ²; + q tan θ²; and y = p tan θ²; + q sec θ²;, then prove that x² - y² = p² - q².
Year of Question:(2014)

Solution:
L.H.S. = x² - y² = (p sec θ²; + q tan θ²;)² - (p tanθ²; + q sec θ²;)²
= p² sec θ²; + q² tan² θ²; + ² pq sec ² tan ² -(p² tan² θ²; + q² sec² θ²; + 2pq sec θ²; tan θ²;)
= p² sec θ²; + 2 tan² _²; + ²pq sec θ²; tan θ²; - p² tan² θ²; - q² sec θ²; - 2pq sec θ²; tan θ²;
= p²(sec² θ²; - tan² θ²;) - q²(sec-² θ²; - tan² θ²;) =
= p² - q² .[sec² θ²; - tan² θ²; = 1
= R.H.S.

Question 20.
Prove the following identity:
Year of Question:(2015)

sin³θ+cos³θ/sinθ+cosθ = 1 - sin θ . cos θ

Solution:

Question 21.
Simplify: 1+tan²A/1+cot²A.
Year of Question:(2014)

Solution:

Question 22.
If x = a cos θ - b sin θ and y = a sin θ + b cos θ, then prove that a² + b² = x² + y².
Year of Question:(2015)

Solution:
R.H.S. = x² + y²
= (a cos θ - b sin θ)² + (a sin θ + b cos θ)²
= a²cos² θ + b² sin² θ - ²ab cos θ sin θ + a² sin² θ + b² cos² θ + ²ab sin θ cos θ
= a²(cos² θ + sin²θ) + b² (sin² θ + cos² θ)
= a² + b² = L.H.S. .[∵ cos² θ + sin² θ = 1

Introduction to Trigonometry Class 10 Important Questions Short Answer - II (3 Marks)

Question 23.
Given 2 cos 3θ = √3, find the value of θ.
Year of Question:(2014)

Solution:
2 cos 3θ = √3 .[Given
cos 3θ = √3/2 ⇒ cos 3θ = cos 30°
30 = 30° ∴ θ = 10°

Question 24.
If cos x = cos 40° . sin 50° + sin 40°. cos 50°, then find the value of x.
Year of Question:(2014)

Solution:
cos x = cos 40° sin 50° + sin 40° cos 50°
cos x = cos 40° sin(90° - 40°) + sin 40°.cos(90° - 40°)
cos x = cos² 40° + sin² 40°
cos x = 1 .[∵ cos² A + sin² A = 1
cos x = cos 0° ⇒ x = 0°

Question 25.
If sin θ = 1/2, then show that 3 cos θ - 4 cos³ θ = 0.
Year of Question:(2014)

Solution:
sin θ = 1/2
sin θ = sin 30° ⇒ θ = 30°
L.H.S = 3 cos θ - 4 cos³ θ
= 3 cos 30° - 4 cos³(30°)

Question 26.
If 5 sin θ = 4, prove that 1/cosθ+1/cotθ = 3
Year of Question:(2013)

Solution:
Given: 5 sin θ = 4

Question 27.

Evaluate: sec 41°. sin 49° + cos 29°.cosec 61°
Year of Question:(2012)

Solution:

Question 28.
Evaluate:
Year of Question:(2012, 2017D)

Solution:

Question 29.
In figure, ∆PQR right angled at Q, PQ = 6 cm and PR = 12 cm. Determine ∠QPR and ∠PRQ.
Year of Question:(2013)

Solution:
In rt. ∆PQR,
PQ² + QR² = PR² .[By Pythogoras‘ theorem

(6)² + QR² = (12)²
QR²= 144 - 36
QR² = 108

Question 30.
Find the value of:
Year of Question:(2013)

Solution:

Question 31.
Prove that: sin²63°+sin²27°/sec²20°-cot²70° + 2 sin 36° sin 42° sec 48° sec 54°
Year of Question:(2017OD)

Solution:

Question 32.
If sin θ = 12/13, 0° <0 < 90°, find the value of: sin²θ-cos²θ/2sin θ.cos θ × 1/tan²θ
Year of Question:(2015)

Solution:

Question 33.
Prove that:
Year of Question:(2012)

Solution:

Question 34.
Prove that: tanθ+secθ-1/tanθ-secθ+1 = 1+sinθ/cosθ
Year of Question:(2012, 2017D)

Solution:

Question 35.
If tan θ = ab, prove that a sinθ-bcosθ/a sinθ+bcosθ = a²?b²/a²+b²
Year of Question:(2013)

Solution:

Question 36.
Prove the identity: (sec A - cos A). (cot A + tan A) = tan A . sec A.
Year of Question:(2014)

Solution:
L.H.S.= (sec A - cos A) (cot A + tan A)

Question 37.
If sec θ + tan θ = p, prove that sin θ = p²?1/p²+1
Year of Question:(2015)

Solution:

Question 38.
Prove that: sinθ-2sin³θ/2cos³θ-cosθ = tan θ
Year of Question:(2015)

Solution:

Question 39.
Prove that: sinθ/1+cosθ + 1+cosθ/sinθ = 2 cosec θ
Year of Question:(2017OD)

Solution:

Introduction to Trigonometry Class 10 Important Questions Long Answer (4 Marks)

Question 40.
In an acute angled triangle ABC, if sin (A + B - C) = 1/2 and cos (B + C - A) = 1/√2, find ∠A, ∠B and ∠C.
Year of Question:(2012)

Solution:

Putting the values of A and B in (iii), we get
67.5° + B + 75o = 180°
B = 180° - 67.5° - 75o = 37.5°
∴ ∠A = 67.5°, ∠B = 37.5° and ∠C = 75°

Question 41.
Evaluate:
Year of Question:(2013)

Solution:

Question 42.
Evaluate the following:
Year of Question:(2015)

Solution:

Question 43.
If θ = 30°, verify the following:
Year of Question:(2014)

(i) cos 3θ = 4 cos3 θ - 3 cos θ
(ii) sin 3θ = 3 sin θ - 4 sin3θ

Solution:

Question 44.
If tan (A + B) = √3 and tan (A - B) = 1/√3 where 0 < A + B < 90°, A > B, find A and B. Also calculate: tan A. sin (A + B) + cos A. tan (A - B).
Year of Question:(2015)

Solution:

Question 45.
Find the value of cos 60° geometrically. Hence find cosec 60°.
Year of Question:(2012, 2017D)

Solution:

Let ∆ABC be an equilateral ∆.
Let each side of triangle be 2a.
Since each angle in an equilateral ∆ is 60°
∴∠A = ∠B = ∠C = 60°
Draw AD ⊥ BC
In ∆ADB and A∆ADC,
AB = AC . [Each = 2a
AD = AD .[Common
∠1 -∠2 . [Each 90°
∴ ∆ADB = ∆ADC .[RHS congruency rule
BD = DC = 2a/2 = a
In rt. ∆ADB, cos 60° = BD/AB = a/2a = 1/2

Question 46.
If tan(20° - 3α) = cot(5α - 20°), then find the value of α and hence evaluate: sin α. sec α . tan α - cosec α . cos α . cot α.
Year of Question:(2014)

Solution:
tan(20° - 3α) = cot(5α - 20°)
tan(20° - 3α) = tan[90° - (5α - 20°)] .[∵ cot θ = tan(90° - θ)]
∴ 20° - 3α= 90° - 5α + 20°
⇒ -3α + 5α = 90° + 20° - 20°
⇒ 2α = 90° ⇒ α = 45°
Now, sin α . sec α tan α- cosec α . cos α . cot α
= sin 45°. sec 45° tan 45° - cosec 45°. cos 45° cot 45°
= 1/√2 × √2 × 1α√2×1/√2 × 1 = 1α1 = 0

Question 47.
If x/a cosθ + y/b sinθ = 1 and x/a sinθ - y/b cosθ = 1, prove that event x²/a² + y²/b² = 2.
Year of Question:(2012, 2017D)

Solution:

Question 48.
If sin θ = c/√c²+d² and d > 0, find the values of cos θ and tan θ. (2013)

Solution:

Question 49.
If cot B = 12/5, prove that tan²B - sin²B = sin⁴ B . sec² B.
Year of Question:(2013)

Solution:
cot B = 12/5 :: AB/BC=12/5

AB = 12k, BC = 5k
In rt. ∆ABC, .[By Pythagoras’ theorem
AC² = AB² + BC²
AC² = (12k)² + (5k)²
AC² = 144k² + 25k²
AC² = 169k²
AC = +13k .[∵ Hypotenuse cannot be -ve

Question 50.
If √3 cot²θ - 4 cot θ + √3 = 0, then find the value of cot² θ + tan²θ.
Year of Question:(2013)

Solution:

Question 51.
Prove that b²x² - a²y² = a²b², if:
Year of Question:(2014)

(i) x = a sec θ, y = b tan θ
(ii) x = a cosec θ, y = b cot θ

Solution:
(i) L.H.S. = b²x² - a²y²
= b²(a sec θ)² - a²(b tan θ)²
= b²a² sec θ - a²b² tan²θ
= b²a²(sec² θ - tan² θ)
= b²a²(1) .[∵ sec²θ - tan² θ = 1
= a²b² = R.H.S.
(ii) L.H.S. = b²x² - a²y²
= b²(a cosec θ)² - a²(b cot θ)²
= b²a² cosec² θ - a²b² cot² θ
= b²a²(cosec²θ - cot² θ)
= b²a² (1) ..[∵ cosec² θ - cot² θ = 1
= a²b²= R.H.S.

Question 52.
If sec θ - tan θ = x, show that sec θ + tan θ = 1/x and hence find the values of cos θ and sin θ.
Year of Question:(2015)

Solution:

Question 53.
If cosec θ + cot θ = p, then prove that cos θ = p²-1/p²+1.
Year of Question:(2012)

Solution:
cosec θ + cot θ = p

Question 54.
If tan θ + sin θ = p; tan θ - sin θ = q; prove that p² - q² = 4√pq.
Year of Question:(2012)

Solution:
L.H.S. = p² - q²
= (tan θ + sin θ)² - (tan θ - sin θ)²
= (tan²θ + sin²θ + 2.tanθ.sinθ) - (tan²θ + sin²θ - 2tan θ sin θ)
= 2 tan θ sin θ+ 2 tan θ sin θ
= 4 tan θ sin θ .(i)

Question 55.
If sin θ + cos θ = m and sec θ + cosec θ = n, then prove that n(m² - 1) = 2m.
Year of Question:(2013)

Solution:
m² - 1 = (sin θ + cos θ)2 - 1
= sin² θ + cos2θ + 2 sin θ cos θ - 1
= 1 + 2 sin θ cos θ - 1
= 2 sin θ cos θ .[sin2 θ + cos² θ = 1
L.H.S. = n(m² - 1)
= (sec θ + cosec θ) 2 sin θ cos θ

Question 56.
Prove that = 2 cosec A
Year of Question:(2012)

Solution:

Question 57.
In ∆ABC, show that sin² A/2 + sin² B+C/2 = 1.
Year of Question:(2013)

Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180° . [Sum of the angles of ∆
∠B + ∠C = 180° - ∠A

Question 58.
Find the value of:
Year of Question:(2013)

Solution:

Question 59.
Prove that: (sin θ + cos θ + 1). (sin θ - 1 + cos θ) . sec θ . cosec θ = 2
Year of Question:(2014)

Solution:
L.H.S. = (sin θ + cos θ + 1) (sin θ - 1 + cos θ) . sec θ cosec θ
= [(sin θ + cos θ) + 1] [(sin θ + cos θ) - 1] . sec θ cosec θ
= [(sin θ + cos θ)2 - (1)²] sec θ cosec θ .[ ∵(a + b)(a - b) = a² - b²
= (sin² θ + cos²θ + 2 sin θ cos θ - 1]. sec θ cosec θ
= (1 + 2 sin θ cos θ - 1). sec θ cosecθ .[ ∵ sin²θ + cos²θ = 1
= (2 sin θ cos θ) 1cosθ/ ? 1/sinθ
2 = R.H.S. .(Hence proved)

Question 60.
Prove that:
Year of Question:(2014)

Solution:

Question 61.
Prove that: (1 + cot A + tan A). (sin A - cos A) = sec³A-csc³A/sec²A.csc²A
Year of Question:(2015)

Solution:

Question 62.
Prove the identity:
Year of Question:(2015)

Solution:

Question 63.
Prove the following trigonometric identities: sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A.
Year of Question:(2015)

Solution:
L.H.S.
= sin A (1 + tan A) + cos A (1 + cot A)

Question 64.
Prove that: (cot A + sec B)² - (tan B - cosec A)² = 2(cot A . sec B + tan B. cosec A)
Year of Question:(2014)

Solution:
L.H.S.
= (cot A + sec B)² - (tan B - cosec A)²
= cot² A + sec² B + 2 cot A sec B - (tan² B + cosec² A - 2 tan B cosec A)
= cot² A + sec² B + 2 cot A sec B - tan²B - cosec²A + 2 tan B cosec A
= (sec²B - tan² B) - (cosec² A - cot² A) + 2(cot A sec B + tan B cosec A)
= 1 - 1 + 2(cot A sec B + tan B cosec A) . [ ∵ sec²B - tan² B = 1
cosec²A - cot² A = 1
= 2(cot A . sec B + tan B . cosec A) = R.H.S.

Question 65.
If x = r sin A cos C, y = r sin A sin C and z = r cos A, then prove that x² + y² + z² = r².
Year of Question:(2017OD)

Solution:
x = r sin A cos C; y = r sin A sin C; z = r cos A
Squaring and adding,
L.H.S. x² + y² + z² = 2 sin² A cos²C + r² sin² A sin² C + r² cos² A
= r² sin² A(cos² C + sin² C) + r² cos² A
= r² sin² A + r² cos² A . [cos²? + sin4? = 1
= r² (sin² A + cos² A) = r² = R.H.S.

Question 66.
Prove that:
Year of Question:(2017OD)

Solution:

Question 67.
In the adjoining figure, ABCD is a rectanlge with breadth BC = 7 cm and ∠CAB = 30°. Find the length of side AB of the rectangle and length of diagonal AC. If the ∠CAB = 60°, then what is the size of the side AB of the rectangle. [Use √3 = 1.73 and √2 = 1.41, if required)
Year of Question:(2014OD)

Solution:

• CBSE Class 10
  SA-1
  Mathematics (Basics)
  Set 1 QP