# Maharashtra Board 10th Class Maths Part 2 Practice Set 6.1 Solutions Chapter 6 Trigonometry

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.

## Practice Set 6.1 Geometry 10th Std Maths Part 2 Answers Chapter 6 Trigonometry

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,
sin2 θ + cos2 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°
∴ AB2 + BC2 = AC2 … [Pythagoras theorem]
∴ (7k)2 + BC2 = (25k)2
∴ 49k2 + BC2 = 625k2
∴ BC2 = 625k2 – 49k2
∴ BC2 = 576k2
∴ 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°
∴ AB2 + BC2 = AC2 …[Pythagoras theorem]
∴ (3k)2 + (4k)2 = AC2
∴ 9k2 + 16k2 = AC2
∴ AC2 = 25k2
∴ 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°
∴ AB2 + BC2 = AC2 … [Pythagoras theorem]
∴ (9k)2 + (40k)2 = AC2
∴ 81k2 + 1600k2 = AC2
∴ AC2 = 1681k2
∴ 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. cos2 θ (1+ tan2 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. sin4 θ – cos4 θ = 1 – 2 cos2 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. = cos2 θ(1 + tan2 θ)
= cos2 θ sec2 θ …[∵ 1 + tan2 θ = sec2 θ]

= 1
= R.H.S.
∴ cos2 θ (1 + tan2 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. = sin4 θ – cos4 θ
= (sin2 θ)2 – (cos2 θ)2
= (sin2 θ + cos2 θ) (sin2 θ – cos2 θ)
= (1) (sin2 θ – cos2 θ) ….[∵ sin2 θ + cos2 θ = 1]
= sin2 θ – cos2 θ
= (1 – cos2 θ) – cos2 θ …[θ sin2 θ = 1 – cos2 θ]
= 1 – 2 cos2 θ
= R.H.S.
∴ sin4 θ – cos4 θ = 1 – 2 cos2 θ

viii. L.H.S. = sec θ + tan θ

xi. L.H.S. = sec4 A (1 – sin4 A) – 2 tan2 A
= sec4 A [12 – (sin2 A)2] – 2 tan2 A
= sec4 A (1 – sin2A) (1 + sin2 A) – 2 tan2 A
= sec4 A cos2A (1 + sin2 A) – 2 tan2A
[ ∵ sin2 θ + cos2 θ = 1 ,∵ 1 – sin2 θ = cos2 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)
sin2 θ + cos2 θ = [______]
Solution:
sin2 θ + cos2 θ = [1]

Question 4.
Write the values of the following trigonometric ratios. (Textbook pg. no. 124)

Solution:

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