Practice Set 6.1 Geometry 10th Standard Maths Part 2 Chapter 6 Trigonometry Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 6 Trigonometry.

10th Standard Maths 2 Practice Set 6.1 Chapter 6 Trigonometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 6.1 Chapter 6 Trigonometry Questions With Answers Maharashtra Board

Question 1.
If sin θ = \(\frac { 7 }{ 25 } \), find the values of cos θ and tan θ.
Solution:
sin θ = \(\frac { 7 }{ 25 } \) … [Given]
We know that,
sin 2 θ + cos 2 θ = 1
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-1
…[Taking square root of both sides] Now, tan θ = \(\frac{\sin \theta}{\cos \theta}\)
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-2
Alternate Method:
sin θ = \(\frac { 7 }{ 25 } \) …(i) [Given]
Consider ∆ABC, where ∠ABC 90° and ∠ACB = θ.
sin θ = \(\frac { AB }{ AC } \) … (ii) [By definition]
∴ \(\frac { AB }{ AC } \) = \(\frac { 7 }{ 25 } \) … [From (i) and (ii)]
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-3
LetAB = 7k and AC = 25k
In ∆ABC, ∠B = 90°
∴ AB 2 + BC 2 = AC 2 … [Pythagoras theorem]
∴ (7k) 2 + BC 2 = (25k) 2
∴ 49k 2 + BC 2 = 625k 2
∴ BC 2 = 625k 2 – 49k 2
∴ BC 2 = 576k 2
∴ BC = 24k …[Taking square root of both sides]
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-3

Question 2.
If tan θ = \(\frac { 3 }{ 4 } \), find the values of sec θ and cos θ.
Solution:
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-4
Alternate Method:
tan θ = \(\frac { 3 }{ 4 } \) …(i)[Given]
Consider ∆ABC, where ∠ABC 90° and ∠ACB = θ.
tan θ = \(\frac { AB }{ BC } \) … (ii) [By definition]
∴ \(\frac { AB }{ BC } \) = \(\frac { 3 }{ 4 } \) … [From (i) and (ii)]
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-5-1
Let AB = 3k and BC 4k
In ∆ABC,∠B = 90°
∴ AB 2 + BC 2 = AC 2 …[Pythagoras theorem]
∴ (3k) 2 + (4k) 2 = AC 2
∴ 9k 2 + 16k 2 = AC 2
∴ AC 2 = 25k 2
∴ AC = 5k …[Taking square root of both sides]
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-6

Question 3.
If cot θ = \(\frac { 40 }{ 9 } \), find the values of cosec θ and sin θ
Solution:
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-7
..[Taking square root of both sides]
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-8
Alternate Method:
cot θ = \(\frac { 40 }{ 9 } \) ….(i) [Given]
Consider ∆ABC, where ∠ABC = 90° and
∠ACB = θ
cot θ = \(\frac { BC }{ AB } \) …(ii) [By defnition]
∴ \(\frac { BC }{ AB } \) = \(\frac { 40 }{ 9 } \) ….. [From (i) and (ii)]
Let BC = 40k and AB = 9k
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-9
In ∆ABC, ∠B = 90°
∴ AB 2 + BC 2 = AC 2 … [Pythagoras theorem]
∴ (9k) 2 + (40k) 2 = AC 2
∴ 81k 2 + 1600k 2 = AC 2
∴ AC 2 = 1681k 2
∴ AC = 41k … [Taking square root of both sides]
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-10

Question 4.
If 5 sec θ – 12 cosec θ = θ, find the values of sec θ, cos θ and sin θ.
Solution:
5 sec θ – 12 cosec θ = 0 …[Given]
∴ 5 sec θ = 12 cosec θ
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-11
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-12

Question 5.
If tan θ = 1, then find the value of
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-13
Solution:
tan θ = 1 … [Given]
We know that, tan 45° = 1
∴ tan θ = tan 45°
∴ θ = 45°
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-14

Question 6.
Prove that:
i. \(\frac{\sin ^{2} \theta}{\cos \theta}+\cos \theta=\sec \theta\)
ii. cos 2 θ (1+ tan 2 θ) = 1
iii. \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sec \theta-\tan \theta\)
iv. (sec θ – cos θ) (cot θ + tan θ) tan θ. sec θ
v. cot θ + tan θ cosec θ. sec θ
vi. \(\frac{1}{\sec \theta-\tan \theta}=\sec \theta+\tan \theta\)
vii. sin 4 θ – cos 4 θ = 1 – 2 cos 2 θ
viii. \(\sec \theta+\tan \theta=\frac{\cos \theta}{1-\sin \theta}\)
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-15
Proof:
i. L.H.S. = \(\frac{\sin ^{2} \theta}{\cos \theta}+\cos \theta\)
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-16

ii. L.H.S. = cos 2 θ(1 + tan 2 θ)
= cos 2 θ sec 2 θ …[∵ 1 + tan 2 θ = sec 2 θ]
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-17
= 1
= R.H.S.
∴ cos 2 θ (1 + tan 2 θ) = 1

Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-18

iv. L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-19
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ. sec θ

v. L.H.S. = cot θ + tan θ
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-20
∴ cot θ + tan θ = cosec θ.sec θ

Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-21

vii. L.H.S. = sin 4 θ – cos 4 θ
= (sin 2 θ) 2 – (cos 2 θ) 2
= (sin 2 θ + cos 2 θ) (sin 2 θ – cos 2 θ)
= (1) (sin 2 θ – cos 2 θ) ….[∵ sin 2 θ + cos 2 θ = 1]
= sin 2 θ – cos 2 θ
= (1 – cos 2 θ) – cos 2 θ …[θ sin 2 θ = 1 – cos 2 θ]
= 1 – 2 cos 2 θ
= R.H.S.
∴ sin 4 θ – cos 4 θ = 1 – 2 cos 2 θ

viii. L.H.S. = sec θ + tan θ
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-22

Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-23
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-24

xi. L.H.S. = sec 4 A (1 – sin 4 A) – 2 tan 2 A
= sec 4 A [1 2 – (sin 2 A) 2 ] – 2 tan 2 A
= sec 4 A (1 – sin 2 A) (1 + sin 2 A) – 2 tan 2 A
= sec 4 A cos 2 A (1 + sin 2 A) – 2 tan 2 A
[ ∵ sin 2 θ + cos 2 θ = 1 ,∵ 1 – sin 2 θ = cos 2 θ]
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-25
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-26

Maharashtra Board Class 10 Maths Chapter 6 Trigonometry Intext Questions and Activities

Question 1.
Fill in the blanks with reference to the figure given below. (Textbook pg. no. 124)
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-27a
Solution:

Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-28

Question 2.
Complete the relations in ratios given below. (Textbook pg, no. 124)
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-29
Solution:
i. \(\frac{\sin \theta}{\cos \theta}\) = [tan θ]
ii. sin θ = cos (90 – θ)
iii. cos θ = (90 – θ)
iv. tan θ × tan (90 – θ) = 1

Question 3.
Complete the equation. (Textbook pg. no, 124)
sin 2 θ + cos 2 θ = [______]
Solution:
sin 2 θ + cos 2 θ = [1]

Question 4.
Write the values of the following trigonometric ratios. (Textbook pg. no. 124)
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-30
Solution:
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-6-Trigonometry-Practice-Set-6.1-31

Maharashtra Board Class 10 Maths Solutions