Class 10 NCERT Solutions Chapter 8 Introduction To Trigonometry Exercise 8.2 (original) (raw)

Last Updated : 13 Aug, 2024

Question 1. Evaluate the following :

****(i) sin 60° cos 30° + sin 30° cos 60°**

**Solution:

Formulas to be used : sin 30° = 1/2

cos 30° = √3/2

sin 60° = 3/2

cos 60° = 1/2

=> (√3/2) * (√3/2) + (1/2) * (1/2)

=> 3/4 +1/4

=> 4 /4

=> 1

****(ii) 2 tan** 2 45° + cos 2 30° – sin 2 60°

**Solution:

Formulas to be used : sin 60° = √3/2

cos 30° = √3/2

tan 45° = 1

=> 2(1)(1) + (√3/2)(√3/2)-(√3/2)(√3/2)

=> 2 + 3/4 - 3/4

=> 2

****(iii) cos 45°/(sec 30°+cosec 30°)**

**Solution:

Formulas to be used : cos 45° = 1/√2

sec 30° = 2/√3

cosec 30° = 2

=> 1/√2 / (2/√3 + 2)

=> 1/√2 / (2+2√3)/√3

=> √3/√2×(2+2 √3) = √3/(2√2+2√6)

=> √3(2√6-2√2)/(2√6+2√2)(2√6-2√2)

=> 2√3(√6-√2) / (2√6)²-(2√2)²

=> 2√3(√6-√2)/(24-8) = 2 √3(√6-√2)/16

=> √3(√6-√2)/8

=> (√18-√6)/8

=> (3√2-√6)/8

****(iv) (sin 30° + tan 45º - cosec 60°)/(sec 30° + cos 60° + cot 45°)**

**Solution:

Formulas to be used : sin 30° = 1/2

tan 45° = 1

cosec 60° = 2/√3

sec 30° = 2/√3

cos 60° = 1/2

cot 45° = 1

=> (1/2+1-2/√3) / (2/√3+1/2+1)

=> (3/2-2/√3)/(3/2+2/√3)

=> (3√3-4/2 √3)/(3√3+4/2 √3)

=> (3√3-4)(3√3-4)/(3√3+4)(3√3-4)

=> (27+16-24√3) / (27-16)

=> (43-24√3)/11

****(v) (5cos** 2 60° + 4sec 2 30° - tan 2 45°)/(sin 2 30° + cos²30°)

**Solution:

Formulas to be used : cos 60° = 1/2

sec 30° = 2/√3

tan 45° = 1

sin 30° = 1/2

cos 30° = √3/2

=> 5(1/2)2+4(2/√3)²-1²/(1/2)+(√3/2)

=> (5/4+16/3-1) / (1/4+3/4)

=> (15+64-12) / 12/(4/4)

=> 67/12

Question 2. Choose the correct option and justify your choice :

****(i) 2tan 30°/1+tan** 2 30° =

****(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°**

****(ii) 1-tan** 2 45°/1+tan 2 45° =

****(A) tan 90° (B) 1 (C) sin 45° (D) 0**

****(iii) sin 2A = 2 sin A is true when A =**

****(A) 0° (B) 30° (C) 45° (D) 60°**

****(iv) 2tan30°/1-tan** 2 30° =

****(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°**

**Solution:

****(i)** In the given equation, substituting the value of tan 30°

As tan 30° = 1/√3

2tan 30°/1+tan230° = 2(1/√3)/1+(1/√3)2

=> (2/√3)/(1+1/3) = (2/√3)/(4/3)

=> 6/4√3 = √3/2

=> sin 60°

The ans is sin 60°.

The correct option is ****(A)**.

****(ii)** In the given equation, substituting the of tan 45°

As tan 45° = 1

1-tan245°/1+tan245° = (1-12)/(1+12)

= 0/2 => 0

The ans is 0.

The correct option is ****(D)**.

****(iii)** sin 2A = 2 sin A is true when A = 0°

sin 2A = sin 0° = 0

2 sin A = 2 sin 0° = 2 × 0 = 0

Another way :

sin 2A = 2sin A cos A

=> 2sin A cos A = 2 sin A

=> 2cos A = 2 => cos A = 1

Now, we have to check which degree value has to be applied, to get the solution as 1.

When 0 degree is applied to cos value we get 1, i.e., cos 0 = 1

Hence, A = 0°

The correct option is ****(A)**.

****(iv)** As tan 30° = 1/√3

2tan30°/1-tan230° = 2(1/√3)/1-(1/√3)2

=> (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°

The correct option is ****(C)**.

Question 3. If tan (A + B) = √3 and tan (A – B) = 1/√3, 0° < A + B ≤ 90°; A > B, find A and B.

**Solution:

tan (A + B) = √3

tan (A + B) = tan 60°

(A + B) = 60° … (i)

tan (A – B) = 1/√3

tan (A – B) = tan 30°

(A – B) = 30° … (ii)

Now add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

A= 45°

Substituting the value of A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

B = 15°

Hence, A = 45° and B = 15°

Question 4. State whether the following are true or false. Justify your answer.

****(i) sin (A + B) = sin A + sin B.**

****(ii) The value of sin θ increases as θ increases.**

****(iii) The value of cos θ increases as θ increases.**

****(iv) sin θ = cos θ for all values of θ.**

****(v) cot A is not defined for A = 0°.**

**Solution:

****(i)** Let us take A = 60° and B = 30°, then

Substitute the values of A and B in the sin (A + B) formula, we get

sin (A + B) = sin (60° + 30°) = sin 90° = 1 and,

sin A + sin B = sin 60° + sin 30°

= √3/2 + 1/2 = (√3 + 1 ) / 2, sin(A + B) ≠ sin A + sin B

Since both the values obtained are not equal.

Hence, the statement is **false.

****(ii)** From the values given below, we can see that as angle(theta) increases value also increases.

sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2 , sin 90° = 1

Thus, the value of sin θ increases as θ increases.

Hence, the statement is **true.

****(iii)** From the values given below, we can see that as angle (theta) increases value decreases.

cos 0° = 1, cos 30° = √3/2 , cos 45° = 1/√2, cos 60° = 1/2, cos 90° = 0

Thus, the value of cos θ decreases as θ increases.

Hence, the statement given above is **false.

****(iv)** sin θ = cos θ, is only true for theta = 45°

Therefore, the above statement is **false.

****(v)** As tan 0° = 0

cot 0° = 1 / tan 0°

= 1 / 0 => undefined

Hence, the given statement is **true.

Summary

Exercise 8.2 in NCERT Class 10 Chapter 8 - Introduction to Trigonometry focuses on trigonometric ratios of complementary angles. It explores the relationships between the trigonometric ratios of an angle and its complement (90° - θ). The exercise emphasizes understanding that sin θ = cos (90° - θ), cos θ = sin (90° - θ), and tan θ = cot (90° - θ). Students learn to apply these relationships to solve problems and simplify trigonometric expressions involving complementary angles.