Here we are providing Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers.

Extra Questions for Class 10 Maths Introduction to Trigonometry with Answers Solutions

Extra Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry with Solutions Answers

Introduction to Trigonometry Class 10 Extra Questions Short Answer Type 1

Question 1.
Find maximum value of \(\frac{1}{\sec \theta}\), 0°≤ θ ≤ 90°.
Solution:
\(\frac{1}{\sec \theta}\), (0° ≤ θ ≤ 90°) (Given)
∵ sec θ is in the denominator
∴ The min. value of sec θ will return max. value for \(\frac{1}{\sec \theta}\).
But the min. value of sec θ is sec 0° = 1.
Hence, the max. value of \(\frac{1}{\sec 0^{\circ}}\) = \(\frac{1}{1}\) = 1

Question 2.
Given that sin θ = \(\frac{a}{b}\), find the value of tan θ.
Solution:
sin θ = \(\frac{a}{b}\)
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 1

Question 3.
If sin θ = cos θ, then find the value of 2 tan θ + cos2 θ.
Solution:
sin θ = cos θ (Given)
It means value of θ = 45°
Now, 2 tan θ + cos2 θ = 2 tan 45° + cos2 45°
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 2

Question 4.
If sin (x – 20)° = cos (3x – 10)°, then find the value of x.
Solution:
sin (x – 20)° = cos (3x – 10)°
⇒ cos [90° – (x – 20)°] = cos (3x – 10)°
By comparing the coefficient
90° – x° + 20° = 3x° – 10° = 110° + 10° = 3x° + x°
120° = 4x°
⇒ \(\frac{120^{\circ}}{4}\) = 30°

Question 5.
If sin2 A = \(\frac{1}{2}\)tan2 45°, where A is an acute angle, then find the value of A.
Solution:
sin2A = \(\frac{1}{2}\)tan2 45°
⇒ sin2A = \(\frac{1}{2}\) (1)2 [∵ tan 45° = 1]
= sin2 A = \(\frac{1}{2}\)
⇒ sin A = \(\frac{1}{\sqrt{2}}\)
Hence, ∠A = 45°

Question 6.
If x = a cos θ, y = b sin θ, then find the value of b2x2 + a2y2 – a2b2.
Solution:
Given x = acos θ, y = b sin θ
b2x2 + a2y2 – a2b2 = b2(acos θ)2 + a2(b sin θ)2 – a2b2
= a2b2 cos2θ + a2b2 sin2 θ – a2b2 = a2b2 (sin2 θ + cos2 θ) – a2b2
= a2b2 – a2b2 = θ (∵ sin2 θ + cos2 θ = 1)

Question 7.
If tan A = cot B, prove that A + B = 90°.
Solution: We have
tan A = cot B
⇒ tan A = tan (90° – B)
A = 90° – B
[∵ Both A and B are acute angles]
⇒ A + B = 90°

Question 8.
If sec A = 2x and tan A = \(\frac{2}{x}\), find the value of 2\(\left(x^{2}-\frac{1}{x^{2}}\right)\) .
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 3

Question 9.
In a ∆ABC, if ∠C = 90°, prove that sin2 A + sin2 B = 1.
Solution:
Since ∠C = 90°
∴ ∠A + ∠B = 180° – ∠C = 90°
Now, sin2 A + sin2 B = sin2 A + sin2 (90° – A) = sin2 A + cos2 A = 1

Question 10.
If sec 4A = cosec (A – 20°) where 4 A is an acute angle, find the value of A.
Solution:
We have
sec 4 A = cosec (A – 20°)
⇒ cosec (90° – 4 A) = cosec (A – 20°)
∴ 90° – 4 A = A – 20°
⇒ 90° + 20° = A + 4 A
⇒ 110° = 5 A
∴ A = \(\frac{110}{5}\) = 22°

Introduction to Trigonometry Class 10 Extra Questions Short Answer Type 2

Question 1.
If sin A = \(\frac{3}{4}\), calculate cos A and tan A.
Solution:
Let us first draw a right ∆ABC in which ∠C = 90°.
Now, we know that
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 4

Question 2.
Given 15 cot A = 8, find sin A and sec A.
Solution:
Let us first draw a right ∆ABC in which ∠B = 90°.
Now, we have, 15 cot A = 8
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 5

Question 3.
In Fig. 10.5, find tan P – cot R.
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 6
Using Pythagoras Theorem, we have
PR2 = PO2 + QR2
⇒ (13)2 = (12)2 + QR2
⇒ 169 = 144 + QR2
⇒ QR2 = 169 – 144 = 25
⇒ QR = 5 cm
Now, tan P = \(\frac{QR}{PQ}\) = \(\frac{5}{12}\) and cot R = \(\frac{QR}{PQ}\) = \(\frac{5}{12}\)
tan P – cot R = \(\frac{5}{12}\) – \(\frac{5}{12}\) = 0

Question 4.
If sin θ + cos θ = √3 , then prove that tan θ + cot θ = 1.
Solution:
sin θ + cos θ = √3
⇒ (sin θ + cos θ)2 = 3
⇒ sin2 θ + cos2 θ + 2 sin θ cos θ = 3
⇒ 2 sin cos θ = 2 (∵ sin2 θ + cos2 θ = 1)
⇒ sin θ. cos θ = 1 = sin2 θ + cos2 θ
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 7
⇒ 1 = tan θ + cot θ = 1
Therefore tan θ + cot θ = 1

Question 5.
Prove that \(\frac { 1-sinθ }{ 1+sinθ } \) = (sec θ – tan θ)2
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 8

Without using tables, evaluate the following (6 to 10).

Question 6.
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 9
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 10

Question 7.
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 11
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 12

Question 8.
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 13
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 14

Question 9.
Evaluate: sin 25° cos 65° + cos 25° sin 65°.
Solution:
sin 25°. cos 65° + cos 25° . sin 65°
= sin (90° – 65°). cos 65° + cos (90° – 65°). sin 65°
= cos 65° . cos 65° + sin 65°. sin 65°
= cos2 65° + sin2 65° = 1.

Question 10.
Without using tables, evaluate the following:
3 cos 68°. cosec 22° – \(\frac{1}{2}\) tan 43°. tan 47°. tan 12°. tan 60°. tan 78°
Solution:
We have,
3 cos 68°. cosec 22° – \(\frac{1}{2}\) tan 43°. tan 47°. tan 12°. tan 60°. tan 78°.
= 3 cos (90° – 22°). cosec 22° – \(\frac{1}{2}\) . {tan 43° . tan (90° – 43°)}. {tan 12°. tan (90° – 12°). tan 60°}
= 3 sin 22°. cosec 22° – \(\frac{1}{2}\)(tan 43° . cot 43°). (tan 12°. cot 12°). tan 60°
= 3 × 1 – × 1 × 1 × √3 = 3 – \(\frac{\sqrt{3}}{2}\) = \(\frac{6-\sqrt{3}}{2}\).

Question 11.
If sin 30 = cos (θ – 6°) where 30 and (θ – 6°) are both acute angles, find the value of θ.
Solution:
According to question:
sin 3θ = cos (θ – 6°)
cos (90° – 30) = cos (θ – 6°) [∵ cos (90° – θ ) = sin θ]
90° – 3θ = θ – 6° [comparing the angles)
= 4θ = 90° + 6° = 96°
θ = \(\frac{96}{4}\) = 24°
Hence, θ = 24°

Question 12.
If sec θ = x + \(\frac{1}{4x}\), prove that sec θ + tan θ = 2x or \(\frac{1}{2x}\).
Solution:
Let sec θ + tan θ = λ …..(i)
We know that, sec2 θ – tan2 θ = 1
(sec θ + tan θ) (sec θ – tan θ) = 1
⇒ λ(sec θ – tan θ) = 1
sec θ – tan θ = \(\frac{1}{\lambda}\) …..(ii)
Adding equations (i) and (ii), we get

Question 13.
Find an acute angle θ, when
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 15
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 16
On comparing we get
⇒ tan θ = √3
⇒ tan θ = tan 60°
= θ = 60°

Question 14.
The altitude AD of a MABC, in which ∠A is an obtuse angle has length 10 cm. If BD = 10 cm and CD = 10√3 cm, determine ∠A.
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 17
∆ABD is a right triangle right angled at D, such that AD = 10 cm and BD = 10 cm.
Let ∠BAD = θ
∴ tan θ = \(\frac{B D}{A D}\)
⇒ tan θ = \(\frac{10}{10}\) = 1
⇒ tan θ = tan 45°
⇒ θ = ∠BAD = 45° … (i)
∆ACD is a right triangle right angled at D such that AD = 10 cm and DC = 10√3 cm.
Let ∠CAD = Φ
∴ tan Φ = \(\frac{CD}{AD}\)
⇒ tan Φ = \(\frac{10 \sqrt{3}}{10}\) = √3
⇒ tan Φ = tan 60°
⇒ Φ = ∠CAD = 60°
From (i) & (ii), we have
∠BAC = ∠BAD + ∠CAD = 45° + 60° = 105°

Question 15.
If cosec θ = \(\frac{13}{12}\), evaluate \(\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}\)
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 18

Question 16.
Prove that
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 19
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 20

Question 17.
Prove that
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 21
Solution:
We have,
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 22

Question 18.
Prove that:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 23
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 24

Question 19.
Prove that:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 25
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 26

Question 20.
Evaluate the following:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 27
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 28
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 29

Question 21.
If tan (A +B) = √3 and tan (A – B) = \(\frac{1}{\sqrt{3}}\); 0° < A + B ≤ 90°; A > B, find A and B.
Solution:
We have, tan (A + B) = √3
⇒ tan (A + B) = tan 60°
∴ A + B = 60° …(i)
Again, tan (A – B) = \(\frac{1}{\sqrt{3}}\)
∴ A – B = 30° … (ii)
Adding (i) and (ii), we have
2A = 90°
⇒ A = 45°
Putting the value of A in (i), we have
45° + B = 60°
∴ B = 60° – 45o = 15°
Hence, A = 45° and B = 15°

Question 22.
If A, B and C are interior angles of a ∆ABC, then show that sin \(\left(\frac{B+C}{2}\right)\) = cos\(\frac{A}{2}\).
Solution:
Since A, B and C are the interior angles of a ∆ABC,
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 30

Question 23.
Prove that: (cosec θ – cot θ)2 = \(\frac {1-\cos\theta}{1+\cos\theta}\)
Solution:
LHS = (cosec θ – cot θ)2
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 31

Question 24.
Prove that:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 32
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 33

Question 25.
Prove that: (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2A.
Solution:
LHS = (sin A + cosec A)2 + (cos A + sec A)2
= sin2 A + coses2 A + 2sin A . cosec A + cos2 A + sec2 A + 2 cos A . sec A
= (sin2 A + coses2 A + 2) + (cos2 A + sec2 A + 2) [sin A. cosec A = 1]
= (sin2 A + cos2 A) + (coses2 A + sec2 A) + 4 [cos A. sec A = 1]
= 1 + 1 + cot2A + 1 + tan2 A + 4
= 7 + tan2 A + cot2 A = RHS [∵ 1 + cot2A = coses2 A and 1 + tan2 A = sec2 A]

Question 26.
Prove that:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 34
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 35
To obtain cot θ in RHS, we have to convert the numerator of LHS in cosine function and denominator in sine function.
Therefore converting sin2 θ = 1 – cos2 θ, we get
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 36

Question 27.
Prove that:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 37
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 38

Question 28.
Prove that:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 39
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 40
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 41

Question 29.
Without using trigonometric tables, prove that:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 42
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 43

Question 30.
Evaluate:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 44
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 45

Introduction to Trigonometry Class 10 Extra Questions Long Answer Type

Question 1.
In ∆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:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 46
We have a right-angled ∆PQR in which ∠Q = 90°.
Let QR = x cm
Therefore, PR = (25 – x) cm
By Pythagoras Theorem, we have
PR2 = PQ2 + QR2
(25 – x)2 = 52 + x2
= (25 – x)2 – x2 = 25
(25 – x – x) (25 – x + x) = 25
(25 – 2x) 25 = 25
25 – 2x = 1
25 – 1 = 2x
= 24 = 2x
∴ x = 12 cm
Hence, QR = 12 cm
PR = (25 – x) cm = 25 – 12 = 13 cm
PQ = 5 cm
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 47

Question 2.
In triangle ABC right-angled at B, if tan A = \(\frac{1}{\sqrt{3}}\) find the value of:
(i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C.
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 48.
We have a right-angled ∆ABC in which ∠B = 90°.
and, tan A = \(\frac{1}{\sqrt{3}}\)
Now, tan A = \(\frac{1}{\sqrt{3}}\) = \(\frac{BC}{AB}\)
Let BC = k and AB = √3k
∴ By Pythagoras Theorem, we have
⇒ AC2 = AB2 + BC2
⇒ AC2 = (√3k)2 + (k)2 = 3k2 + k2
⇒ AC2 = 4k2
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 49

Question 3.
If cot θ = \(\frac{7}{8}\), evaluate:
(i)
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 50
(ii) cot2 θ
Solution:
Let us draw a right triangle ABC in which ∠B = 90° and ∠C = θ.
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 51
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 52

Question 4.
If 3 cot A = 4, check whether \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos2 A – sin2 A or not.
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 53
Let us consider a right triangle ABC in which ∠B = 90°
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 54
Let AB = 4k and BC = 3k
∴ By Pythagoras Theorem
AC2 = AB2 + BC2
AC = (4k)2 + (3k)2 = 16k2 + 9k2
AC2 = 25k2
∴ AC = 5k
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 55

Question 5.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 56
Let us consider a right-angled ∆ABC in which ∠B = 90°.
For ∠A we have
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 57

Question 6.
Prove that
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 58
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 59

Question 7.
Prove that:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 60
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 61

Question 8.
Prove that
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 62
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 63
= 2 cosec2 A tan2 A = 2(1 + cot2 A). tan2 A
= 2 tan2 A + 2 tan2 A. cot2 A (∵ tan A cot A = 1)
= 2 + 2 tan2 A = 2(1 + tan2 A) = 2 sec2 A = RHS.

Question 9.
Prove that: (sin θ + sec θ)2 + (cos θ + cosec θ)2 = (1 + sec θ cosec θ)2.
Solution:
LHS = (sin θ + sec θ)2 + (cos θ + cosec θ)2
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 64
= (1 + sec θ cosec θ)2 = RHS.

Question 10.
Prove that:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 65
Solution:
In order to show that,
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 66
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 67

Question 11.
Prove that: \(\frac { cosecθ\quad +\quad cotθ }{ cosecθ\quad -\quad cotθ } \) = (cosec θ + cot θ) = 1 + 2 cot2 θ + 2 cosec θ cot θ.
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 68
= cosec2 θ + cot2 θ + 2 cosec θ. cot θ
= (1 + cot2 θ) + cot2 θ + 2 cosec θ. cot θ
= 1 + 2 cot2 θ + 2 cosec . cot θ = RHS.

Question 12.
Prove that: 2 sec θ – sec θ – 2 cosec θ + cosec θ = cot – tan θ.
Solution:
LHS = 2 sec θ – sec θ – 2 cosec2 θ + cosec θ
= 2 (sec2 θ) – (sec2 θ)2 – 2 (cosec2 θ) + (cosec θ)2
= 2 (1 + tan2 θ) – (1 + tan4 θ)2 – 2(1 + cot2 θ) + (1 + cot2 θ)4
= 2 + 2 tan2 θ – (1 + 2 tan2 θ + tan2 θ) – 2 – 2 cot2 θ + (1 + 2 cot2 θ + cot θ) =
= 2 + 2 tan2 θ – 1 – 2 tan2 θ – tan4 θ – 2 – 2 cot2 θ + 1 + 2 cot4 θ + cot4 θ
= cot4 θ – tan4 θ = RHS

Question 13.
Prove that: (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\).
Solution:
LHS = (cosec A – sin A) (sec A – cos A)
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 69
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 70

Introduction to Trigonometry Class 10 Extra Questions HOTS

Question 1.
Prove that:
\(\frac{\tan \theta}{1-\cot \theta}\) + \(\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ = 1 + tan θ + cot θ.
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 71

For second part
Now from (i), we have
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 72

Question 2.
If tan A = n tan B and sin A = m sin B, prove that cos2 A = \(\frac{m^{2}-1}{n^{2}-1}\)
Solution:
We have to find cos2 A in terms of m and n. This means that the angle B is to be eliminated from
the given relations.
Now, tan A = n tan B
⇒ tan B = \(\frac{1}{n}\) tan A
⇒ cot B = \(\frac{n}{\tan A}\)
and
sin A = m sin B
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 73

Question 3.
Prove the following identity, where the angle involved is acute angle for which the expressions are defined.
\(\frac { cosA-sinA+1 }{ cos A + sin A-1 } \) = cosec A + cot A, using the identity cosec2 A = 1 + cot2 A.
Solution:
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 74
= cosec A + cot A = RHS.

Question 4.
If x sin3 + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, prove x2 + y2 = 1.
Solution:
We have, x sin3 + y cos3 θ = sin θ cos θ
⇒(x sin θ) sin2 θ + (y cos θ) cos2 θ sin θ cos θ
⇒ x sin θ (sin2 θ) + (x sin θ) cos2 = sin θ cos [∵ x sin θ = y cos θ]
⇒ x sin θ (sin2 θ + cos2 θ) = sin θ cos θ
⇒ x sin θ = sin θ cos
⇒ x = cos θ
Now, we have x sin θ = y cos θ
⇒ cos o sin θ = y cos θ [∵ x = cos θ]
⇒ y = sin θ
Hence, x2 + y2 = cos2 θ + sin2 θ = 1.

Question 5.
If tan θ + sin θ = m and tan θ – sin θ = n, show that (m2 – n2) = 4√mn.
Solution:
We have, given tan θ + sin θ = m, and tan θ – sin θ = n, then
LHS = (m2 – n2),= (tan θ + sin θ)2 – (tan θ – sin )2
= tan2 θ + sin2 θ + 2 tan θ sin θ – tan2 θ – sin2 θ + 2 tan θ sin θ
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 75

Question 6.
If cosec θ – sin θ = l and sec θ – cos θ = m, prove that l2 m2 (l2 + m2 + 3) = 1.
Solution:
LHS = l2 m2 (l2 + m2 + 3).
= (cosec θ – sin θ)2 (sec θ – cos θ)2 {(cosec θ – sin θ)2 + (sec θ – cos θ)2 + 3}
Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers 76