Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry
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. (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. (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 = \(\frac{117^{\circ}}{3}\) = 39°
Question 3.
If tan α = \(\sqrt{3}\) and tan β = \(\frac{1}{\sqrt{3}}\),0 < α, β < 90°, find the value of cot (α + β). (2012)
Solution:
tan α = \(\sqrt{3}\) = tan 60° …(i)
tan β = \(\frac{1}{\sqrt{3}}\) = 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. (2012, 2017D)
Solution:
sin θ – cos θ = 0 = sin θ = cos θ
⇒ \(\frac{\sin \theta}{\cos \theta}\) = 1 ⇒ tan θ = 1 ⇒ θ = 45°
Now, sin4θ + cos4θ
= sin4 45° + cos4 45°
= \(\left(\frac{1}{\sqrt{2}}\right)^{4}+\left(\frac{1}{\sqrt{2}}\right)^{4}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\)
Question 5.
If sec θ + tan θ = 7, then evaluate sec θ – tan θ. (2017OD)
Solution:
We know that,
sec2θ – tan2θ = 1
(sec θ + tan θ) (sec θ – tan θ) = 1
(7) (sec θ – tan θ) = 1 …[sec θ + tan θ = 7; (Given)
∴ sec θ – tan θ = \(\frac{1}{7}\)
Question 6.
Evaluate: 10. \(\frac{1-\cot ^{2} 45^{\circ}}{1+\sin ^{2} 90^{\circ}}\). (2014)
Solution:
Question 7.
If cosec θ = \(\frac{5}{4}\), find the value of cot θ. (2014)
Solution:
We know that, cot2θ = cosec2θ – 1
= \(\left(\frac{5}{4}\right)^{2}\) – 1 ⇒ \(\frac{25}{16}\) – 1 ⇒ \(\frac{25-16}{16}\)
coť2θ = \(\frac{9}{16}\) i cot θ = \(\frac{3}{4}\)
Question 8.
If θ = 45°, then what is the value of 2 sec2θ + 3 cosec2θ ? (2014)
Solution:
2 sec2θ + 3 cosec2θ = 2 sec2 45° + 3 cosec2 45°
= 2(\(\sqrt{2}\))2 + 3 (\(\sqrt{2}\))2 = 4 + 6 = 10
Question 9.
If \(\sqrt{3}\) sin θ = cos θ, find the value of \(\frac{3 \cos ^{2} \theta+2 \cos \theta}{3 \cos \theta+2}\). (2015)
Solution:
\(\sqrt{3}\) sin θ = cos θ … [Given
Question 10.
Evaluate: sin2 19° + sin771°. (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°? (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 = \(\frac{a}{x}\), find the value of \(\frac{x}{\sqrt{a^{2}+x^{2}}}\). (2013)
Solution:
Question 13.
If in a right angled ∆ABC, tan B = \(\frac{12}{5}\), then find sin B. (2014)
Solution:
1st method:
tan B = \(\frac{12}{5}\) ∴ cot B = \(\frac{5}{12}\)
cosec2 B = 1 + cot2 B
= 1 + \([\left(\frac{5}{12}\right)^{2}/latex] = 1 + [latex]\)
= \(\frac{144+25}{144}=\frac{169}{144}\)
cosec B = \(\frac{13}{12}\) ∴ sin B = \(\frac{12}{13}\)
2nd method:
tan B = \(\frac{12}{5}\)
tan B = \(\frac{AC}{BC}\)
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 = \(\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{12 \mathrm{k}}{13 \mathrm{k}}=\frac{12}{13}\)
Question 14.
If ∆ABC is right angled at B, what is the value of sin (A + C). (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°. (2013)
Solution:
tan 15°. tan 25°, tan 60°. tan 65°. tan 75° – tan 30°
= tan(90° – 75°) tan(90° – 65°). \(\sqrt{3}\) . tan 65°. tan 75° – \(\frac{1}{\sqrt{3}}\)
Question 16.
Express cot 75° + cosec 75° in terms of trigonometric ratios of angles between 0° and 30°. (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. (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 \(\left(\frac{A+B}{2}\right)\) = cos\(\left(\frac{c}{2}\right)\). (2012)
Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180° …(Angle sum property of ∆
∠A + ∠B = 180° – ∠C
Question 19.
If x = p sec θ + q tan θ and y = p tan θ + q sec θ, then prove that x2 – y2 = p2 – q2. (2014)
Solution:
L.H.S. = x2 – y2
= (p sec θ + q tan θ)2 – (p tan θ + q sec θ)2
= p2 sec θ + q2 tan2 θ + 2 pq sec 2 tan 2 -(p2 tan2 θ + q2 sec2 θ + 2pq sec θ tan θ)
= p2 sec θ + 2 tan2 θ + 2pq sec θ tan θ – p2 tan2 θ – q2 sec θ – 2pq sec θ tan θ
= p2(sec2 θ – tan2 θ) – q2(sec?2 θ – tan2 θ) =
= p2 – q2 …[sec2 θ – tan2 θ = 1
= R.H.S.
Question 20.
Prove the following identity: (2015)
\(\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta+\cos \theta}\) = 1 – sin θ . cos θ
Solution:
Question 21.
Simplify: \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\). (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. (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 = \(\sqrt{3}\), find the value of θ. (2014)
Solution:
2 cos 3θ = \(\sqrt{3}\) …[Given
cos 3θ = \(\frac{\sqrt{3}}{2}\) ⇒ 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. (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 = \(\frac{1}{2}\), then show that 3 cos θ – 4 cos3 θ = 0. (2014)
Solution:
sin θ = \(\frac{1}{2}\)
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 \(\frac{1}{\cos \theta}+\frac{1}{\cot \theta}\) = 3 (2013
Solution:
Given: 5 sin θ = 4
Question 27.
Evaluate: sec 41°. sin 49° + cos 29°.cosec 61° (2012)
Solution:
Question 28.
Evaluate: (2012, 2017D)
Solution:
Question 29.
In figure, ∆PQR right angled at Q, PQ = 6 cm and PR = 12 cm. Determine ∠QPR and ∠PRQ. (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: (2013)
Solution:
Question 31.
Prove that: \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\sec ^{2} 20^{\circ}-\cot ^{2} 70^{\circ}}\) + 2 sin 36° sin 42° sec 48° sec 54° (2017OD)
Solution:
Question 32.
If sin θ = \(\frac{12}{13}\), 0° <0 < 90°, find the value of: \(\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cdot \cos \theta} \times \frac{1}{\tan ^{2} \theta}\) (2015)
Solution:
Question 33.
Prove that: (2012)
Solution:
Question 34.
Prove that: \(\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}=\frac{1+\sin \theta}{\cos \theta}\) (2012, 2017D)
Solution:
Question 35.
If tan θ = \(\frac{a}{b}\), prove that \(\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\) (2013)
Solution:
Question 36.
Prove the identity: (sec A – cos A). (cot A + tan A) = tan A . sec A. (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 = \(\frac{p^{2}-1}{p^{2}+1}\) (2015)
Solution:
Question 38.
Prove that: \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\) = tan θ (2015)
Solution:
Question 39.
Prove that: \(\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}\) = 2 cosec θ (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) = \(\frac{1}{2}\) and cos (B + C – A) = \(\frac{1}{\sqrt{2}}\), find ∠A, ∠B and ∠C. (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: (2013)
Solution:
Question 42.
Evaluate the following: (2015)
Solution:
Question 43.
If θ = 30°, verify the following: (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) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\) where 0 < A + B < 90°, A > B, find A and B. Also calculate: tan A. sin (A + B) + cos A. tan (A – B). (2015)
Solution:
Question 45.
Find the value of cos 60° geometrically. Hence find cosec 60°. (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 = \(\frac{2 a}{2}\) = a
In rt. ∆ADB, cos 60° = \(\frac{\mathrm{BD}}{\mathrm{AB}}=\frac{a}{2 a}=\frac{1}{2}\)
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. (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°
= \(\frac{1}{\sqrt{2}} \times \sqrt{2} \times 1-\sqrt{2} \times \frac{1}{\sqrt{2}} \times 1=1-1=0\)
Question 47.
If \(\frac{x}{a}\)cosθ + \(\frac{y}{b}\)sinθ = 1 and \(\frac{x}{a}\)sinθ – \(\frac{y}{b}\) cosθ = 1, prove that event \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 2. (2012, 2017D)
Solution:
Question 48.
If sin θ = \(\frac{c}{\sqrt{c^{2}+d^{2}}}\) and d > 0, find the values of cos θ and tan θ. (2013)
Solution:
Question 49.
If cot B = \(\frac{12}{5}\), prove that tan2B – sin2B = sin4 B . sec2 B. (2013)
Solution:
cot B = \(\frac{12}{5}\) :: \(\frac{A B}{B C}=\frac{12}{5}\)
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 \(\sqrt{3}\) cot2θ – 4 cot θ + \(\sqrt{3}\) = 0, then find the value of cot2 θ + tan2θ. (2013)
Solution:
Question 51.
Prove that b2x2 – a2y2 = a2b2, if: (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 = \(\frac{1}{x}\) and hence find the values of cos θ and sin θ. (2015)
Solution:
Question 53.
If cosec θ + cot θ = p, then prove that cos θ = \(\frac{p^{2}-1}{p^{2}+1}\). (2012)
Solution:
cosec θ + cot θ = p
Question 54.
If tan θ + sin θ = p; tan θ – sin θ = q; prove that p2 – q2 = \(4 \sqrt{p q}\). (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. (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 (2012)
Solution:
Question 57.
In ∆ABC, show that sin2 \(\frac{\mathbf{A}}{2}\) + sin2 \(\frac{\mathbf{B}+\mathbf{C}}{\mathbf{2}}\) = 1. (2013)
Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180° … [Sum of the angles of ∆
∠B + ∠C = 180° – ∠A
Question 58.
Find the value of: (2013)
Solution:
Question 59.
Prove that: (sin θ + cos θ + 1). (sin θ – 1 + cos θ) . sec θ . cosec θ = 2 (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). \(\frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta}\)
= 2 = R.H.S. …(Hence proved)
Question 60.
Prove that: (2014)
Solution:
Question 61.
Prove that: (1 + cot A + tan A). (sin A – cos A) = \(\frac{\sec ^{3} A-\csc ^{3} A}{\sec ^{2} A \cdot \csc ^{2} A}\) (2015)
Solution:
Question 62.
Prove the identity: (2015)
Solution:
Question 63.
Prove the following trigonometric identities: sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A. (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) (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 – tan2 B – cosec2 A + 2 tan B cosec A
= (sec2 B – 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. (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 … [cos2θ + sin2θ = 1
= r2 (sin2 A + cos2 A) = r2 = R.H.S.
Question 66.
Prove that: (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 \(\sqrt{3}\) = 1.73 and \(\sqrt{2}\) = 1.41, if required) (2014OD)
Solution: