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 tan²45° + cos²30° – sin²60°

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) (5cos²60° + 4sec² 30° – tan²45°)/(sin²30° + 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+tan²30° =

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

(ii) 1-tan²45°/1+tan²45° =

(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-tan²30° =

(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.