# Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry – Exercise 8.2

### 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°

^{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°)

^{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+tan

^{2}30° = 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-tan

^{2}45°/1+tan^{2}45° = (1-1^{2})/(1+1^{2})= 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/√32tan30°/1-tan

^{2}30° = 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°, thenSubstitute 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° = 0cot 0° = 1 / tan 0°

= 1 / 0 => undefined

Hence, the given statement is

true.

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