# NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Exercise – 8.2) ## Ex 8.2 Question 1.

Evaluate the following:

(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan2 45° + cos2 30° – sin2 60 Solution :

(i) sin 60° cos 30° + sin 30° cos 60°

sin 30° = 1/2

cos 30° = √3/2

sin 60° = √3/2

cos 60°= 1/2

putting the values of

sin 60° cos 30° + sin 30° cos 60°

= √3/2 ×√3/2 + (½) ×(½ ) = 3/4+1/4

= 4/4

=1

(ii) 2 tan2 45° + cos2 30° – sin2 60

sin 60° = √3/2

cos 30° = √3/2

tan 45° = 1

putting the values of

2 tan2 45° + cos2 30° – sin2 60

= 2(1)+ (√3/2)2-(√3/2)2

= 2 + 3/4 – 3/4

= 2

(iii) cos 45°/(sec 30°+cosec 30°)

cos 45° = 1/√2

sec 30° = 2/√3

cosec 30° = 2

putting the values of multiply numerator and denominator by √2,  sin 30° = 1/2

tan 45° = 1

cosec 60° = 2/√3

sec 30° = 2/√3

cos 60° = 1/2

cot 45° = 1

putting the values of  cos 60° = 1/2

sec 30° = 2/√3

tan 45° = 1

sin 30° = 1/2

cos 30° = √3/2

(5cos260° + 4sec230° – tan245°)/(sin30° + cos30°)

= 5(½)2+4(2/√3)2-12/(½)2+(√3/2)2

= (5/4+16/3-1)/(¼+¾)

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

= 67/12

## Ex 8.2 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) Putting the value of tan 30°

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°

Hence , Correct answer is (A).

(ii) Putting the value of tan 45°

tan 45° = 1

1-tan245°/1+tan245° = (1-12)/(1+12)

= 0/2 = 0

Hence , Correct answer is (D).

(iii) Sin 0° = 0

Hence , Correct answer is (A).

(iv) Putting the value of tan 30°

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°

Hence , Correct answer is (C).

## Ex 8.2 Question 3.

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

Solution: ## Ex 8.2 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°.

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