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# NCERT Exemplar - Introduction to Trigonometry Class 10 Notes | EduRev

## Class 10 : NCERT Exemplar - Introduction to Trigonometry Class 10 Notes | EduRev

The document NCERT Exemplar - Introduction to Trigonometry Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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Exercise 8.1 Page No: 89

Choose the correct answer from the given four options:
1. If cos A = 4/5, then the value of tan A is
(a) 3/5
(b) 3/4
(c) 4/3
(d) 5/3
Ans: (b)
Explanation:
According to the question,
cos A = 4/5 …(1)
We know,
tan A = sinA/cosA
To find the value of sin A,
We have the equation,
sin2 θ +cos2 θ = 1
So, sin θ = √ (1-cos2 θ)
Then,
sin A = √ (1-cos2 A) …(2)
sin2 A = 1-cos2 A
sin A = √(1-cos2 A)
Substituting equation (1) in (2),
We get,
Sin A = √(1-(4/5)2)
= √(1-(16/25))
= √(9/25)
= ¾
Therefore, 2. If sin A = 1/2, then the value of cot A is
(a) √3
(b) 1/√3
(c) √3/2
(d) 1

Ans: (a)
Explanation:
(A) √3
According to the question,
Sin A = ½ … (1)
We know that, To find the value of cos A.
We have the equation,
sin2 θ +cos2 θ =1
So, cos θ = √(1-sin2 θ)
Then,
cos A = √(1-sin2 A) … (3)
cos2 A = 1-sin2 A
cos A = √ (1 - sin2 A)
Substituting equation 1 in 3, we get,
cos A = √(1 - 1/4) = √(3/4) = √3/2
Substituting values of sin A and cos A in equation 2, we get
cot A = (√3/2) × 2 = √3

3. The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is
(a) – 1
(b) 0
(c) 1
(d) 3 2
Ans: (b)
Explanation:
According to the question,
We have to find the value of the equation,
cosec(75° + θ) – sec(15° - θ) – tan(55° + θ) + cot(35° - θ)
= cosec[90°-(15°-θ)] – sec(15°-θ) – tan(55°+θ) + cot[90°-(55°+θ)]
Since, cosec (90°- θ) = sec θ
And, cot(90°-θ) = tan θ
We get,
= sec(15° - θ) – sec(15° - θ) – tan(55° + θ) + tan(55° + θ)
= 0

4. Given that sinθ = a b , then cosθ is equal to
(a) b/√(b– a2
(b) b/a
(c) √(b- a2)/b
(d) a/√(b- a2)
Ans: (c)
Explanation:
According to the question,
sin θ = a/b
We know, sin2 θ +cos2 θ = 1
sin2 A = 1 - cos2 A
sin A = √(1-cos2 A
So, cos θ = √(1 - a2/b2 ) = √((b- a2)/b2 ) = √(b- a2 )/b
Hence, cos θ = √(b2 – a2 )/b

5. If cos (α + β) = 0, then sin (α – β) can be reduced to
(a) cos β
(b) cos 2β
(c) sin α
(d) sin 2α
Ans: (b)
Explanation:
According to the question,
cos(α+β) = 0
Since, cos 90° = 0
We can write,
cos(α+β)= cos 90°
By comparing cosine equation on L.H.S and R.H.S,
We get,
(α + β) = 90°
α = 90°- β
Now we need to reduce sin (α - β),
So, we take,
sin(α - β) = sin(90°- β - β) = sin(90°- 2β)
sin(90° - θ) = cos θ
So, sin(90° - 2β) = cos 2β
Therefore, sin(α - β) = cos 2β

6. The value of (tan1° tan2° tan3° … tan89°) is
(a) 0
(b) 1
(c) 2
(d) 1/2
Ans: (b)
Solution:
tan 1°. tan 2°.tan 3° …… tan 89°
= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.tan 45°.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°
Since, tan 45° = 1,
= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°
= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan(90°-44°).tan(90°-43°)…tan(90°-3°). tan(90°-2°).tan(90°-1°)
Since, tan(90°-θ) = cot θ,
= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.cot 44°.cot 43°…cot 3°.cot 2°.cot 1°
Since, tan θ = (1/cot θ)
= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1. (1/tan 44o). (1/tan 43o)… (1/tan 3o). (1/tan 2o). (1/tan 1o) = 1
Hence, tan 1°.tan 2°.tan 3° …… tan 89° = 1

7. If cos 9α = sinα and 9α < 90°, then the value of tan5α is
(a) 1/√3
(b) √3
(c) 1
(d) 0
Ans: (c)
Solution:
According to the question,
cos 9∝ = sin ∝ and 9∝<90°
i.e. 9α is an acute angle
We know that,
sin(90°-θ) = cos θ
So,
cos 9∝ = sin (90°- ∝)
Since, cos 9∝ = sin(90°- 9∝) and sin(90°- ∝) = sin∝
Thus, sin (90°- 9∝) = sin∝
90°- 9∝ = ∝
10∝ = 90°
∝ = 9°
Substituting ∝ = 9° in tan 5∝, we get,
tan 5∝ = tan (5 × 9) = tan 45° = 1
∴, tan 5∝ = 1

Exercise 8.2 Page No: 93

Write ‘True’ or ‘False’ and justify your answer in each of the following:
1. tan 47o/cot 43° = 1
Solution:
True
Justification:
Since, tan (90° - θ) = cot θ 2. The value of the expression (cos223° – sin267°) is positive.
Solution:
False
Justification:
Since, (a- b2) = (a + b)(a - b)
cos2 23° – sin2 67° =(cos 23°+sin 67°)(cos 23°-sin 67°)
= [cos 23°+sin(90°-23°)] [cos 23°-sin(90°-23°)]
= (cos 23°+cos 23°)(cos 23°-cos 23°) (∵sin(90°-θ) = cos θ)
= (cos 23°+cos 23°).0
= 0, which is neither positive nor negative

3. The value of the expression (sin 80° – cos 80°) is negative.
Solution:
False
Justification:
We know that,
sin θ increases when 0° ≤ θ ≤ 90°
cos θ decreases when 0° ≤ θ ≤ 90°
And (sin 80°-cos 80°) = (increasing value-decreasing value)
= a positive value.
Therefore, (sin 80°-cos 80°) > 0.

4. √((1– cos2θ) sec2 θ)= tan θ
Solution:
True
Justification: 5. If cosA + cos2A = 1, then sin2A + sin4A = 1.
Solution:
True
Justification:
According to the question,
cos A+cos2 A = 1
i.e., cos A = 1- cos2 A
Since,
sin2 θ+cos2 θ = 1
sin2 θ = 1- cos2 θ)
We get,
cos A = sin2 A …(1)
Squaring L.H.S and R.H.S,
cos2 A = sin4 A …(2)
To find sin2A+sin4 A = 1
Adding equations (1) and (2),
We get
sin2A + sin4 A= cos A + cos2 A
Therefore, sin2A + sin4 A = 1

6. (tan θ + 2) (2 tan θ + 1) = 5 tan θ + sec2 θ.
Solution:
False
Justification:
L.H.S = (tan θ + 2) (2 tan θ + 1)
= 2 tan2 θ + tan θ + 4 tan θ + 2
= 2 tan2θ + 5 tan θ + 2
Since, sec2 θ – tan2 θ = 1, we get, tan2θ = sec2 θ - 1
= 2(sec2 θ-1) +5 tan θ + 2
= 2 sec2 θ - 2 + 5 tan θ + 2
= 5 tan θ + 2 sec2 θ ≠ R.H.S
∴, L.H.S ≠ R.H.S

Exercise 8.3 Page No: 95

Prove the following (from Q.1 to Q.7):
1. sin θ/(1 + cos θ) + (1 + cos θ)/sin θ = 2cosec θ
Solution:
L.H.S = R.H.S
Hence proved.

2. tan A/(1 + secA) – tan A/(1 - secA) = 2cosec A
Solution:

L.H.S: Since,
sec2A – tan2A = 1
sec2A – 1 = tan2A = R.H.S
Hence proved.

3. If tan A = ¾, then sinA cosA = 12/25
Solution:
According to the question,
tan A = ¾
We know,
tan A = perpendicular/ base
So,
tan A = 3k/4k
Where,
Perpendicular = 3k
Base = 4k Using Pythagoras Theorem,
(hypotenuse)2 = (perpendicular)2 + (base)2
(hypotenuse)2 = (3k)2+ (4k)2 = 9k2+16k2 = 25k2
hypotenuse = 5k
To find sin A and cos A, Hence, proved.

4. (sin α + cos α) (tan α + cot α) = sec α + cosec α
Solution:

L.H.S:
(sin α + cos α) (tan α + cot α)
As we know, = R.H.S
Hence, proved.

5. (√3+1) (3 – cot 30°) = tan60° – 2 sin 60°
Solution:
L.H.S: (√3 + 1) (3 – cot 30°)
= (√3 + 1) (3 – √3) [∵cos 30° = √3]
= (√3 + 1) √3 (√3 – 1) [∵(3 – √3) = √3 (√3 – 1)]
= ((√3)2– 1) √3 [∵ (√3+1)(√3-1) = ((√3)2 – 1)]
= (3-1) √3
= 2√3
Similarly solving R.H.S: tan3 60° – 2 sin 60°
Since, tan 60o = √3 and sin 60o = √3/2,
We get,
(√3)3 – 2.(√3/2) = 3√3 – √3
= 2√3
Therefore, L.H.S = R.H.S
Hence, proved.

6. 1 + (cot2 α/1+cosec α) = cosec α
Solution: And, we know that, 7. tan θ + tan (90° – θ) = sec θ sec (90° – θ)
Solution:
L.H.S =
Since, tan (90° – θ) = cot θ
tan θ + tan (90° – θ)= tan θ + cot θ Exercise 8.4 Page No: 99

1. If cosecθ + cotθ = p, then prove that cosθ = (p2 – 1)/ (p2 + 1).
Solution:
According to the question,
cosec θ + cot θ = p
Since, Hence, proved.

2. Prove that √(sec2 θ + cosec2 θ) = tan θ + cot θ
Solution:
L.H.S =
√ (sec2 θ + cosec2 θ)
Since, = R.H.S
Hence, proved.

3. The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.
Solution:
Let PR = h meter, be the height of the tower.
The observer is standing at point Q such that, the distance between the observer and tower is QR = (20 + x) m, where
QR = QS + SR = 20 + x
∠PQR = 30°
∠ PSR = θ In ∆PQR, Rearranging the terms,
We get 20 + x = √3h
⇒ x = √3h – 20 …eq.1
In ∆PSR,
tan θ = h/x
Since, angle of elevation increases by 15o when the observer moves 20 m towards the tower.
We have,
θ = 30° + 15° = 45°
So,
tan 45o = h/x
⇒ 1 = h/x
⇒ h = x
Substituting x=h in eq. 1, we get
h = √3 h – 20
⇒ √3 h – h = 20
⇒ h (√3 – 1) = 20 = 10 (√3 + 1)
Hence, the required height of the tower is 10 (√3 + 1) meter.

4. If 1 + sin2θ = 3sinθ cosθ , then prove that tanθ = 1 or 1/2.
Solution:
Given: 1+sin2 θ = 3 sin θ cos θ
Dividing L.H.S and R.H.S equations with sin2 θ,
We get, cosec2 θ + 1 = 3 cot θ
Since,
cosec2 θ – cot2 θ = 1 ⇒ cosec2 θ = cot2 θ + 1
⇒ cot2 θ + 1 + 1 = 3 cot θ
⇒ cot2 θ +2 = 3 cot θ
⇒ cot2 θ –3 cot θ +2 = 0
Splitting the middle term and then solving the equation,
⇒ cot2 θ – cot θ –2 cot θ +2 = 0
⇒ cot θ(cot θ -1)–2(cot θ +1) = 0
⇒ (cot θ – 1)(cot θ – 2) = 0
⇒ cot θ = 1, 2
Since,
tan θ = 1/cot θ
tan θ = 1, ½
Hence, proved.

5. Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
Solution:
Given: sin θ + 2 cos θ = 1
Squaring on both sides,
(sin θ +2 cos θ)2 = 1
⇒ sin2 θ + 4 cos2 θ + 4sin θcos θ = 1
Since, sin2 θ = 1 – cos2 θ and cos2 θ = 1 – sin2 θ
⇒ (1 – cos2 θ) + 4(1 – sin2 θ) + 4sin θcos θ = 1
⇒ 1 – cos2 θ + 4 – 4 sin2 θ + 4sin θcos θ = 1
⇒ – 4 sin2 θ – cos2 θ + 4sin θcos θ = – 4
⇒ 4 sin2 θ + cos2 θ – 4sin θcos θ = 4
We know that,
a2+ b2 – 2ab = ( a – b)2
So, we get,
(2sin θ – cos θ)2 = 4
⇒ 2sin θ – cos θ = 2
Hence proved.

6. The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √st .
Solution:
Let BC = s; PC = t
Let height of the tower be AB = h.
∠ABC = θ and ∠APC = 90° – θ
(∵ the angle of elevation of the top of the tower from two points P and B are complementary)  ⇒ h2 = st
⇒ h = √st
Hence the height of the tower is √st.

7. The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.
Solution:
Let SQ = h be the tower.
∠SPQ = 30° and ∠SRQ = 60°
According to the question, the length of shadow is 50 m long hen angle of elevation of the sun is 30° than when it was 60°. So,
PR = 50 m and RQ = x m  ⇒ 50√3+h = 3h
⇒ 50√3 = 3h – h
⇒ 3h – h = 50√3
⇒ 2h = 50√3
⇒ h = (50√3)/2
⇒ h = 25√3
Hence, the required height is 25√3 m.

8. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are α and β, respectively. Prove that the height of the tower is [h tan α/(tan β – tan α)].
Solution:

Given that a vertical flag staff of height h is surmounted on a vertical tower of height H(say),
such that FP = h and FO = H.
The angle of elevation of the bottom and top of the flag staff on the plane is ∠PRO = α and ∠FRO = β respectively In ∆PRO, we have Hence, proved.

9. If tanθ + secθ = l, then prove that secθ = (l2 + 1)/2l.
Solution:
Given: tan θ+ sec θ = l …eq. 1
Multiplying and dividing by (sec θ – tan θ) on numerator and denominator of L.H.S, So, sec θ – tan θ = 1 …eq.2
Adding eq. 1and eq. 2, we get
(tan θ + sec θ) + (sec θ – tan θ) = 1 Hence, proved.

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## Mathematics (Maths) Class 10

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