Chapter 8 Introduction to Trigonometry and its Equation NCERT Exemplar Solutions Exercise 8.1 Class 10 Maths

Chapter 8 Introduction to Trigonometry and its Equation NCERT Exemplar Solutions Exercise 8.1 Class 10 Maths

Chapter Name

NCERT Maths Exemplar Solutions for Chapter 8 Introduction to Trigonometry and its Equation Exercise 8.1

Book Name

NCERT Exemplar for Class 10 Maths

Other Exercises

  • Exercise 8.2
  • Exercise 8.3
  • Exercise 8.4

Related Study

NCERT Solutions for Class 10 Maths

Exercise 8.1 Solutions

Multiple Choice Questions

Choose the correct answer from the given four options:

1. If cos A = 4/3,  then the value of tan A is 
(A) 3/5
(B) 3/4 
(C) 4/3
(D) 5/3

Solution

As given in the question, 
cos A = 4/3
Also,
tan A = sin A/cos A
We have,
sin2 θ + cos2 θ = 1
So,

Putting equation (1) in (2), 
We get, 


2. If sin A = 1/2, then the value of cot A is 
(A) √3 
(B) 1/√3
(C) √3/2
(D) 1 

Solution

As given in the question, 
sin A = 1/2 
We have, 
cot A = cos A/sin A 
Now to find the value of cos A,  
We have, 
sin2 θ + cos2 θ = 1 
So, 

= √3/2 
Putting values of sin A and cos A in equation 2, 
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) 32

Solution

As given in the question,
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)]
As,
cosec (90°- θ) = sec θ
Also,
cot(90°-θ) = tan θ
So,
cosec[90°-(15°-θ)] – sec(15°-θ) – tan(55°+θ) + cot[90°-(55°+θ)]
= sec(15°-θ) – sec(15°-θ) – tan(55°+θ) + tan(55°+θ)
= 0


4. Given that sin θ = a/b, then cos θ is equal to 
(A) b/√(b2 - a2)
(B) b/a 
(C) √(b2 - a2)/b
(D) a/(√(b2 - a2)
Solution
As given in the question, 
sin θ = a/b
We have, 
sin2 θ + cos2 θ = 1
⇒ sin2 A = 1 - cos2 A

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
Solution
As given in the question, 
cos (α + β) = 0 
We have, 
cos 90° = 0
Also, we can write,
cos(α+β)= cos 90°
On 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)
Also,
sin(90°-θ) = cos θ
So,
sin(90°-2β) = cos 2β
Hence, sin(α-β) = cos 2β

6. The value of (tan1° tan2° tan3° ... tan89°) is
(A) 0
(B) 1
(C) 2
(D) 1/2
Solution
According to question, 
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°
Also,
tan 45° = 1,
= tan1°.tan 2°.tan 3°...tan 43°.tan 44°.1.tan 46°.tan 47°...tan 87°.tan 88°.tan 89°
Now, we can write,
= 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°)
As,
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°
And also,
tan θ = (1/cotθ)
= tan1°.tan 2°.tan 3°...tan 43°.tan 44°.1.  (1/tan 44°).(1/tan43°)....(1/tan2°).(1/tan 1°). = 1 
Therefore, tan 1°. tan2° . tan 3° ....... tan 89° = 1

7. If cos 9α = sinα and 9α < 90°, then the value of tan 5α is 
(A) 1/√3 
(B) √3
(C) 1 
(D) 0
Solution
As given in the question, 
cos 9α = sinα
and 
9α < 90° 
Which means, 
9α is an acute angle 
We know that,
sin(90°-θ) = cos θ
So,
cos 9∝ = sin (90°-∝)
Also,
cos 9∝ = sin(90°-9∝)
and
sin(90°-∝) = sin∝
So,
sin (90°-9∝) = sin∝
⇒ 90°-9∝ =∝
⇒ 10∝ = 90°
⇒ ∝ = 9°
Putting ∝ = 9° in tan 5∝,
we get,
tan 5∝ = tan (5×9)
= tan 45°
= 1
So,
tan 5∝ = 1

8. If  ΔABC is right angled at C, then the value of cos(A + B) is 
(A) 0 
(B) 1 
(C) 1/2 
(D) √3/2
Solution
(A) 
We have, in ∆ABC,
Sum of three angles = 180°
∠A + ∠B + ∠C = 180°
Also the triangle is right angled at C so,
∠C = 90°
∠A + ∠B + 90° = 180°
⇒ A + B = 90°
cos(A + B) = cos90°
= 0

9. If sin A + sin2 A = 1, then the value of the expression (cos2 A + cos4 A) is
(A) 1
(B) 1/2 
(C) 2 
(D) 3 
Solution
(A) 
We have, 
sin A + sin2 A = 1 
⇒ sin A = 1 - sin2 A
= cos2 A  [as, sin2 θ + cos2 θ = 1]
sin A = cos2 A 
On squaring both sides, we get 
sin2 A = cos4 A
⇒ 1 - cos2 A = cos4 A
⇒ cos2 A + cos4 A = 1  

10. Given that sin ∝ = 1/2 and cos ∝ = 1/2, then the value of ( ∝ + β) is 
(A) 0°
(B) 30°
(C) 60°
(D) 90°
Solution
Sin ∝ = 1/2 = sin 30°
so,
∝ = 30° 
and cos β = 1/2 
= cos 60° 
⇒ β = 60° 
⇒ (∝ + β) = 30° + 60° = 90°

11. The value of the expression is [(sin2 22° + sin2 68)°)/(cos2 22° + cos2 68°) + sin2 63° + cos 63°sin27°] is 
(A) 3 
(B) 2 
(C) 1
(D) 0 
Solution

12.  If 4tan θ = 3, then [(4sin θ - cos θ)/(4 sinθ + cosθ) is equal to 
(A) 2/3 
(B) 1/3 
(C) 1/2 
(D) 3/4
Solution
(C) 
We have, 
4 tan θ = 3 
tan θ = 3/4 
We have, 

13. If sin θ - cos θ = 0, then the value of (sin4 θ + cos4 θ) is 
(A)1 
(B) 3/4 
(C) 1/2 
(D) 1/4 
Solution
We have, 
sin θ - cos θ = 0 
⇒ sin θ = cos θ
⇒ sin θ/cos θ = 1 
⇒ tan θ = 1 
also, 
tan  45° = 1 
⇒ tan θ = tan  45° 
⇒ θ = 45°
So, 
(sin4 θ + cos4 θ) = (sin4 45° + cos4 45°) 
= 1/2 

14. sin (45° + θ) - cos (45° - θ) is equal to 
(A) 2cos θ
(B) 0
(C) 2sinθ
(D) 1
Solution
(B) 
sin(45° + θ) - cos(45° - θ) = cos[90° - (45° + θ) - cos(45° - θ)
[as, cos(90° - θ) = sinθ]
= cos(45° - θ) - cos(45° - θ)
= 0

15. A pole 6 m high casts a shadow 2√3 m long on the ground, then the Sun's elevation is 
(A) 60°
(B) 45°
(C) 30°
(D) 90°
Solution
Taking BC = 6 m be the height of the pole and AB = 2√3 m be the length of the shadow on the ground. 
And let the Sun's elevation be θ.
In ABC, 
tan θ = BC/AC 
⇒ tan θ = 6/2√3
⇒ tan θ = √3 = tan 60°
⇒ tan θ = tan 60°
⇒ θ = 60°
So, the Sun's elevation is 60°.
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