## Maharashtra State Board Class 10 Maths Solutions Chapter 6 Trigonometry Practice Set 6.1

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: