# 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 8.2Exercise 8.3Exercise 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 Î¸ + cos2 Î¸ = 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 Î¸ + cos2 Î¸ = 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° + Î¸) – sec (15° – Î¸) – tan (55° + Î¸) + cot (35° – Î¸)] is
(A) –1
(B) 0
(C) 1
(D) 32

Solution

As given in the question,
The value of the equation,
cosec(75°+Î¸) – sec(15°-Î¸) – tan(55°+Î¸) + cot(35°-Î¸)
= cosec[90°-(15°-Î¸)] – sec(15°-Î¸) – tan(55°+Î¸) + cot[90°-(55°+Î¸)]
As,
cosec (90°- Î¸) = sec Î¸
Also,
cot(90°-Î¸) = tan Î¸
So,
cosec[90°-(15°-Î¸)] – sec(15°-Î¸) – tan(55°+Î¸) + cot[90°-(55°+Î¸)]
= sec(15°-Î¸) – sec(15°-Î¸) – tan(55°+Î¸) + tan(55°+Î¸)
= 0

4. Given that sin Î¸ = a/b, then cos Î¸ 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 Î¸ = a/b
We have,
sin2 Î¸ + cos2 Î¸ = 1
⇒ sin2 A = 1 - cos2 A

5. If cos (Î± + Î²) = 0, then sin (Î± – Î²) can be reduced to
(A) cos Î²
(B) cos 2Î²
(C) sin Î±
(D) sin 2Î±
Solution
As given in the question,
cos (Î± + Î²) = 0
We have,
cos 90° = 0
Also, we can write,
cos(Î±+Î²)= cos 90°
On comparing cosine equation on L.H.S and R.H.S,
We get,
(Î±+Î²)= 90°
Î± = 90°-Î²
Now we need to reduce sin (Î± -Î² ),
So, we take,
sin(Î±-Î²) = sin(90°-Î²-Î²)
= sin(90°-2Î²)
Also,
sin(90°-Î¸) = cos Î¸
So,
sin(90°-2Î²) = cos 2Î²
Hence, sin(Î±-Î²) = cos 2Î²

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°-Î¸) = cot Î¸,
= tan1°.tan 2°.tan 3°...tan 43°.tan 44°.1.cot 44°.cot 43°...cot 3°.cot 2°.cot 1°
And also,
tan Î¸ = (1/cotÎ¸)
= 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 9Î± = sinÎ± and 9Î± < 90°, then the value of tan 5Î± is
(A) 1/√3
(B) √3
(C) 1
(D) 0
Solution
As given in the question,
cos 9Î± = sinÎ±
and
9Î± < 90°
Which means,
9Î± is an acute angle
We know that,
sin(90°-Î¸) = cos Î¸
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  Î”ABC 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 Î¸ + cos2 Î¸ = 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 ( ∝ + Î²) is
(A) 0°
(B) 30°
(C) 60°
(D) 90°
Solution
Sin ∝ = 1/2 = sin 30°
so,
∝ = 30°
and cos Î² = 1/2
= cos 60°
⇒ Î² = 60°
⇒ (∝ + Î²) = 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 Î¸ = 3, then [(4sin Î¸ - cos Î¸)/(4 sinÎ¸ + cosÎ¸) is equal to
(A) 2/3
(B) 1/3
(C) 1/2
(D) 3/4
Solution
(C)
We have,
4 tan Î¸ = 3
tan Î¸ = 3/4
We have,

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

14. sin (45° + Î¸) - cos (45° - Î¸) is equal to
(A) 2cos Î¸
(B) 0
(C) 2sinÎ¸
(D) 1
Solution
(B)
sin(45° + Î¸) - cos(45° - Î¸) = cos[90° - (45° + Î¸) - cos(45° - Î¸)
[as, cos(90° - Î¸) = sinÎ¸]
= cos(45° - Î¸) - cos(45° - Î¸)
= 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 Î¸.
In ABC,
tan Î¸ = BC/AC
⇒ tan Î¸ = 6/2√3
⇒ tan Î¸ = √3 = tan 60°
⇒ tan Î¸ = tan 60°
⇒ Î¸ = 60°
So, the Sun's elevation is 60°.