Std 10 Mathematics Part 2 · Chapter 6

6. Trigonometry

Solutions to Practice Sets 6.1, 6.2 and Problem Set 6, with proofs of identities and figures for heights and distances. Maharashtra State Board (Balbharati), Std 10 Mathematics Part 2 (Geometry).

Practice Set 6.1

1If sinθ = 725, find the values of cosθ and tanθ.

sin²θ + cos²θ = 1

∴ (725)² + cos²θ = 1

∴ cos²θ = 1 − 49625 = 576625

∴ cosθ = 2425

tanθ = sinθcosθ = 725 × 2524 = 724

∴ cosθ = 2425 and tanθ = 724

2If tanθ = 34, find the values of secθ and cosθ.

1 + tan²θ = sec²θ

∴ sec²θ = 1 + 916 = 2516

∴ secθ = 54

cosθ = 1secθ = 45

∴ secθ = 54 and cosθ = 45

3If cotθ = 409, find the values of cosecθ and sinθ.

1 + cot²θ = cosec²θ

∴ cosec²θ = 1 + 160081 = 168181

∴ cosecθ = 419

sinθ = 1cosecθ = 941

∴ cosecθ = 419 and sinθ = 941

4If 5secθ − 12cosecθ = 0, find the values of secθ, cosθ and sinθ.

5secθ = 12cosecθ

∴ secθcosecθ = 125

But secθcosecθ = 1cosθ × sinθ = sinθcosθ = tanθ

∴ tanθ = 125

sec²θ = 1 + tan²θ = 1 + 14425 = 16925

∴ secθ = 135

cosθ = 1secθ = 513

sinθ = tanθ × cosθ = 125 × 513 = 1213

∴ secθ = 135, cosθ = 513 and sinθ = 1213

5If tanθ = 1 then, find the values of sinθ + cosθsecθ + cosecθ.

tanθ = 1 = tan 45°, so θ = 45°.

sin 45° = cos 45° = 1√2 and sec 45° = cosec 45° = √2

sinθ + cosθsecθ + cosecθ = 1√2 + 1√2√2 + √2 = 2√22√2 = √22√2 = 12

∴ The value is 12

6Prove that:

(1) sin²θcosθ + cosθ = secθ

Proof: L.H.S. = sin²θcosθ + cosθ

= sin²θ + cos²θcosθ

= 1cosθ … sin²θ + cos²θ = 1

= secθ = R.H.S.

(2) cos²θ(1 + tan²θ) = 1

Proof: L.H.S. = cos²θ(1 + tan²θ)

= cos²θ × sec²θ … 1 + tan²θ = sec²θ

= cos²θ × 1cos²θ

= 1 = R.H.S.

(3) √(1 − sinθ1 + sinθ) = secθ − tanθ

Proof: L.H.S. = √(1 − sinθ1 + sinθ)

= √((1 − sinθ)(1 − sinθ)(1 + sinθ)(1 − sinθ)) … multiplying by (1 − sinθ)

= √((1 − sinθ)²1 − sin²θ) = √((1 − sinθ)²cos²θ)

= 1 − sinθcosθ = 1cosθ − sinθcosθ

= secθ − tanθ = R.H.S.

(4) (secθ − cosθ)(cotθ + tanθ) = tanθ secθ

Proof: L.H.S. = (secθ − cosθ)(cotθ + tanθ)

= (1cosθ − cosθ)(cosθsinθ + sinθcosθ)

= 1 − cos²θcosθ × cos²θ + sin²θsinθ cosθ

= sin²θcosθ × 1sinθ cosθ = sinθcos²θ

= sinθcosθ × 1cosθ = tanθ secθ = R.H.S.

(5) cotθ + tanθ = cosecθ secθ

Proof: L.H.S. = cotθ + tanθ = cosθsinθ + sinθcosθ

= cos²θ + sin²θsinθ cosθ = 1sinθ cosθ

= 1sinθ × 1cosθ = cosecθ secθ = R.H.S.

(6) 1secθ − tanθ = secθ + tanθ

Proof: L.H.S. = 1secθ − tanθ

= secθ + tanθ(secθ − tanθ)(secθ + tanθ)

= secθ + tanθsec²θ − tan²θ

= secθ + tanθ1 … sec²θ − tan²θ = 1

= secθ + tanθ = R.H.S.

(7) sin⁴θ − cos⁴θ = 1 − 2cos²θ

Proof: L.H.S. = sin⁴θ − cos⁴θ

= (sin²θ + cos²θ)(sin²θ − cos²θ)

= 1 × (sin²θ − cos²θ)

= (1 − cos²θ) − cos²θ

= 1 − 2cos²θ = R.H.S.

(8) secθ + tanθ = cosθ1 − sinθ

Proof: L.H.S. = secθ + tanθ = 1cosθ + sinθcosθ = 1 + sinθcosθ

= (1 + sinθ)(1 − sinθ)cosθ(1 − sinθ)

= 1 − sin²θcosθ(1 − sinθ) = cos²θcosθ(1 − sinθ)

= cosθ1 − sinθ = R.H.S.

(9) If tanθ + 1tanθ = 2, then show that tan²θ + 1tan²θ = 2

Proof: tanθ + 1tanθ = 2

Squaring both sides, (tanθ + 1tanθ)² = 4

∴ tan²θ + 2 × tanθ × 1tanθ + 1tan²θ = 4

∴ tan²θ + 2 + 1tan²θ = 4

∴ tan²θ + 1tan²θ = 2

(10) tan A(1 + tan²A)² + cot A(1 + cot²A)² = sin A cos A

Proof: L.H.S. = tan A(sec²A)² + cot A(cosec²A)²

= tan A × cos⁴A + cot A × sin⁴A

= sin Acos A × cos⁴A + cos Asin A × sin⁴A

= sin A cos³A + cos A sin³A

= sin A cos A (cos²A + sin²A)

= sin A cos A = R.H.S.

(11) sec⁴A (1 − sin⁴A) − 2tan²A = 1

Proof: L.H.S. = sec⁴A (1 − sin²A)(1 + sin²A) − 2tan²A

= sec⁴A × cos²A (1 + sin²A) − 2tan²A

= sec²A (1 + sin²A) − 2tan²A … sec²A × cos²A = 1

= sec²A + sin²Acos²A − 2tan²A

= sec²A + tan²A − 2tan²A

= sec²A − tan²A = 1 = R.H.S.

(12) tanθsecθ − 1 = tanθ + secθ + 1tanθ + secθ − 1

Proof: L.H.S. = tanθsecθ − 1 = tanθ(secθ + 1)(secθ − 1)(secθ + 1)

= tanθ(secθ + 1)sec²θ − 1 = tanθ(secθ + 1)tan²θ

= secθ + 1tanθ …… (I)

Now, (secθ + 1)(tanθ + secθ − 1) = tanθ(secθ + 1) + (secθ + 1)(secθ − 1)

= tanθ(secθ + 1) + sec²θ − 1

= tanθ(secθ + 1) + tan²θ

= tanθ(tanθ + secθ + 1)

∴ secθ + 1tanθ = tanθ + secθ + 1tanθ + secθ − 1 …… (II)

∴ L.H.S. = R.H.S. … from (I) and (II)

Practice Set 6.2

1A person is standing at a distance of 80 m from a church looking at its top. The angle of elevation is of 45°. Find the height of the church.

AB is the church, C is the person.

In right angled Δ ABC, ∠ ACB = 45°.

tan 45° = ABBC

∴ 1 = h80

∴ h = 80

∴ The height of the church is 80 m

2From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60°. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. (√3 = 1.73)

AB is the lighthouse, C is the ship. AD is horizontal.

∠ DAC = 60° is the angle of depression.

∠ ACB = ∠ DAC = 60° … alternate angles, AD ∥ BC

In right angled Δ ABC, tan 60° = ABBC

∴ √3 = 90x

∴ x = 90√3 = 90√33 = 30√3

= 30 × 1.73 = 51.9

∴ The ship is 51.9 m from the lighthouse.

3Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building?

AB is the first building and CD is the second. AE ∥ BD.

□ ABDE is a rectangle, so AE = BD = 12 m and ED = AB = 10 m.

In right angled Δ AEC, tan 60° = CEAE

∴ √3 = x12

∴ x = 12√3

CD = CE + ED = 12√3 + 10

= 12 × 1.73 + 10 = 20.76 + 10 = 30.76 … taking √3 = 1.73

∴ The height of the second building is (10 + 12√3) m, that is about 30.76 m

4Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops is 22 metre. Find the angle made by the wire with the horizontal.

AB and CD are the poles and AC is the wire. CE ∥ DB.

□ EBDC is a rectangle, so EB = CD = 7 m.

AE = AB − EB = 18 − 7 = 11 m

In right angled Δ AEC, sin θ = AEAC = 1122 = 12

∴ θ = 30°

∴ The wire makes an angle of 30° with the horizontal.

5A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.

BC is the standing part and CA is the broken part.

Height of the tree = BC + CA = x + y.

In right angled Δ CBA, tan 60° = BCBA

∴ √3 = x20

∴ x = 20√3

cos 60° = BACA

∴ 12 = 20y

∴ y = 40

Height = 20√3 + 40 = 20 × 1.73 + 40 = 34.6 + 40 = 74.6 … taking √3 = 1.73

∴ The height of the tree was (40 + 20√3) m, that is about 74.6 m

6A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60° with the ground. Assuming that the string is straight, find the length of the string. (√3 = 1.73)

A is the kite and CA is the string.

In right angled Δ ABC, sin 60° = ABAC

∴ √32 = 60x

∴ x = 120√3 = 120√33 = 40√3

= 40 × 1.73 = 69.2

∴ The length of the string is 69.2 m

Problem Set 6

1Choose the correct alternative answer for the following questions.

(1) sinθ cosecθ = ? — (A) 1
sinθ × 1sinθ = 1
(2) cosec 45° = ? — (B) √2
cosec 45° = 1sin 45° = √2
(3) 1 + tan²θ = ? — (C) sec²θ
This is a basic trigonometric identity.
(4) When we see at a higher level, from the horizontal line, angle formed is …… — (A) angle of elevation.
Looking up from the horizontal gives an angle of elevation.

2If sinθ = 1161, find the values of cosθ using trigonometric identity.

sin²θ + cos²θ = 1

∴ cos²θ = 1 − 1213721 = 36003721

∴ cosθ = 6061

3If tanθ = 2, find the values of other trigonometric ratios.

cotθ = 1tanθ = 12

sec²θ = 1 + tan²θ = 1 + 4 = 5, so secθ = √5

cosθ = 1secθ = 1√5

sinθ = tanθ × cosθ = 2 × 1√5 = 2√5

cosecθ = 1sinθ = √52

∴ sinθ = 2√5, cosθ = 1√5, cotθ = 12, secθ = √5, cosecθ = √52

4If secθ = 1312, find the values of other trigonometric ratios.

cosθ = 1secθ = 1213

tan²θ = sec²θ − 1 = 169144 − 1 = 25144, so tanθ = 512

sinθ = tanθ × cosθ = 512 × 1213 = 513

cosecθ = 135 and cotθ = 125

∴ sinθ = 513, cosθ = 1213, tanθ = 512, cosecθ = 135, cotθ = 125

5Prove the following.

(1) secθ (1 − sinθ) (secθ + tanθ) = 1

Proof: L.H.S. = 1cosθ (1 − sinθ) (1cosθ + sinθcosθ)

= 1cosθ (1 − sinθ) × 1 + sinθcosθ

= 1 − sin²θcos²θ = cos²θcos²θ

= 1 = R.H.S.

(2) (secθ + tanθ) (1 − sinθ) = cosθ

Proof: L.H.S. = (1cosθ + sinθcosθ)(1 − sinθ)

= (1 + sinθ)(1 − sinθ)cosθ

= 1 − sin²θcosθ = cos²θcosθ

= cosθ = R.H.S.

(3) sec²θ + cosec²θ = sec²θ × cosec²θ

Proof: L.H.S. = 1cos²θ + 1sin²θ

= sin²θ + cos²θcos²θ × sin²θ

= 1cos²θ × sin²θ

= 1cos²θ × 1sin²θ = sec²θ × cosec²θ = R.H.S.

(4) cot²θ − tan²θ = cosec²θ − sec²θ

Proof: L.H.S. = cot²θ − tan²θ

= (cosec²θ − 1) − (sec²θ − 1) … 1 + cot²θ = cosec²θ, 1 + tan²θ = sec²θ

= cosec²θ − sec²θ = R.H.S.

(5) tan⁴θ + tan²θ = sec⁴θ − sec²θ

Proof: L.H.S. = tan²θ(tan²θ + 1)

= (sec²θ − 1) × sec²θ

= sec⁴θ − sec²θ = R.H.S.

(6) 11 − sinθ + 11 + sinθ = 2sec²θ

Proof: L.H.S. = (1 + sinθ) + (1 − sinθ)(1 − sinθ)(1 + sinθ)

= 21 − sin²θ = 2cos²θ

= 2sec²θ = R.H.S.

(7) sec⁶x − tan⁶x = 1 + 3sec²x × tan²x

Proof: L.H.S. = (sec²x)³ − (tan²x)³

Using a³ − b³ = (a − b)³ + 3ab(a − b),

= (sec²x − tan²x)³ + 3sec²x tan²x (sec²x − tan²x)

= 1³ + 3sec²x tan²x × 1 … sec²x − tan²x = 1

= 1 + 3sec²x × tan²x = R.H.S.

(8) tanθsecθ + 1 = secθ − 1tanθ

Proof: L.H.S. = tanθsecθ + 1 = tanθ(secθ − 1)(secθ + 1)(secθ − 1)

= tanθ(secθ − 1)sec²θ − 1 = tanθ(secθ − 1)tan²θ

= secθ − 1tanθ = R.H.S.

(9) tan³θ − 1tanθ − 1 = sec²θ + tanθ

Proof: L.H.S. = (tanθ − 1)(tan²θ + tanθ + 1)tanθ − 1 … a³ − b³ = (a − b)(a² + ab + b²)

= tan²θ + tanθ + 1

= (1 + tan²θ) + tanθ

= sec²θ + tanθ = R.H.S.

(10) sinθ − cosθ + 1sinθ + cosθ − 1 = 1secθ − tanθ

Proof: L.H.S. = tanθ − 1 + secθtanθ + 1 − secθ … dividing numerator and denominator by cosθ

In the numerator, write 1 = sec²θ − tan²θ:

Numerator = (tanθ + secθ) − (sec²θ − tan²θ)

= (secθ + tanθ) − (secθ + tanθ)(secθ − tanθ)

= (secθ + tanθ)(1 − secθ + tanθ)

∴ L.H.S. = (secθ + tanθ)(tanθ + 1 − secθ)tanθ + 1 − secθ = secθ + tanθ

= (secθ + tanθ)(secθ − tanθ)secθ − tanθ = sec²θ − tan²θsecθ − tanθ

= 1secθ − tanθ = R.H.S.

6A boy standing at a distance of 48 meters from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.

AB is the building and C is the boy.

In right angled Δ ABC, tan 30° = ABBC

∴ 1√3 = h48

∴ h = 48√3 = 48√33 = 16√3

= 16 × 1.73 = 27.68 … taking √3 = 1.73

∴ The height of the building is 16√3 m, that is about 27.68 m

7From the top of the light house, an observer looks at a ship and finds the angle of depression to be 30°. If the height of the light-house is 100 meters, then find how far the ship is from the light-house.

AB is the lighthouse and C is the ship. AD is horizontal.

∠ ACB = ∠ DAC = 30° … alternate angles, AD ∥ BC

In right angled Δ ABC, tan 30° = ABBC

∴ 1√3 = 100x

∴ x = 100√3

= 100 × 1.73 = 173 … taking √3 = 1.73

∴ The ship is 100√3 m, that is about 173 m, from the lighthouse.

8Two buildings are in front of each other on a road of width 15 meters. From the top of the first building, having a height of 12 meter, the angle of elevation of the top of the second building is 30°. What is the height of the second building?

AB is the first building and CD is the second. AE ∥ BD.

□ ABDE is a rectangle, so AE = BD = 15 m and ED = AB = 12 m.

In right angled Δ AEC, tan 30° = CEAE

∴ 1√3 = x15

∴ x = 15√3 = 5√3

CD = CE + ED = 5√3 + 12

= 5 × 1.73 + 12 = 8.65 + 12 = 20.65 … taking √3 = 1.73

∴ The height of the second building is (12 + 5√3) m, that is about 20.65 m

9A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20 m. If the platform is 2 m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)

P is the foot of the ladder on the platform and L is its top.

In right angled Δ LMP, sin 70° = LMPL

∴ 0.94 = h20

∴ h = 20 × 0.94 = 18.8

Height from the ground = LM + MN = 18.8 + 2 = 20.8

∴ The ladder can reach a height of 20.8 m

10★While landing at an airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing. (sin 20° = 0.342)

A is the plane when landing starts and C is where it touches the ground.

Distance travelled = speed × time = 200 × 543600 = 3 km

∴ AC = 3 km = 3000 m

∠ ACB = ∠ DAC = 20° … alternate angles

In right angled Δ ABC, sin 20° = ABAC

∴ 0.342 = h3000

∴ h = 3000 × 0.342 = 1026

∴ The plane was at a height of 1026 m (1.026 km).

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