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
…[Taking square root of both sides] Now, tan θ = \(\frac{\sin \theta}{\cos \theta}\)
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)]
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]
Question 2.
If tan θ = \(\frac { 3 }{ 4 } \), find the values of sec θ and cos θ.
Solution:
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)]
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]
Question 3.
If cot θ = \(\frac { 40 }{ 9 } \), find the values of cosec θ and sin θ
Solution:
..[Taking square root of both sides]
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
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]
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
Question 5.
If tan θ = 1, then find the value of
Solution:
tan θ = 1 … [Given]
We know that, tan 45° = 1
∴ tan θ = tan 45°
∴ θ = 45°
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. \(\sec \theta+\tan \theta=\frac{\cos \theta}{1-\sin \theta}\)
Proof:
i. L.H.S. = \(\frac{\sin ^{2} \theta}{\cos \theta}+\cos \theta\)
ii. L.H.S. = cos
2
θ(1 + tan
2
θ)
= cos
2
θ sec
2
θ …[∵ 1 + tan
2
θ = sec
2
θ]
= 1
= R.H.S.
∴ cos
2
θ (1 + tan
2
θ) = 1
iv. L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ. sec θ
v. L.H.S. = cot θ + tan θ
∴ cot θ + tan θ = cosec θ.sec θ
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
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 Chapter 6 Trigonometry Intext Questions and Activities
Question 1.
Fill in the blanks with reference to the figure given below. (Textbook pg. no. 124)
Solution:
Question 2.
Complete the relations in ratios given below. (Textbook pg, no. 124)
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)
Solution: