Introduction to Trigonometry
Question 1.
Evaluate :
(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°
= √3/2 x √3/2 + 1/2 x 1/2
= 3/4 + 1/4
= (3+1)/4
= 4/4
= 1.
(ii) 2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + (√3/2)2 – (√3/2)2
= 2 + 3/4 – 3/4
= 2






Question 2.
Choose the correct option and justify:
(i) (2tan30∘)/(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) O
(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) none of these
Solution:
(i) (A)

(ii) (D)
Because (1−tan245∘)/(1+tan245∘) = (1−1)/(1+1) = 0/2 = 0
(iii) (A)
Because when A – 0°, sin 2A = sin 0° = 0
and, 2 sin A = 2 sin 0°= 2 x 0 = 0
or sin 2 A = 2 sin A, when A = 0°
(iv) (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 means tan (A + B) = tan 60°
So, A + B = 60° … (1)
tan (A – B) = 1/√3 means tan (A – B) = tan 30°
So, A – B = 30° … (2)
Solving (1) and (2), we get
A = 45° and 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 0 = cos 0 for all values of 0.
(v) cot A is not defined for A = 0°.
Solution:
(i) False. Because
When A = 60° and B = 30°. Then,
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
So, sin (A + B) ≠ sin A + sin B
(ii) True. Because, it is clear from the table below :

that the value of sin θ increases as θ increases.
(iii) False. Because it is clear from the table below :

that the value of cos θ decreases as θ increases.
(iv) False. Because it is only true for θ = 45°.
(sin 45° = 1/√2 = cos 45°)
(v) True. Because tan 0° = 0 and
cot 0° = 1/tan0∘ = 1/0, i.e., not defined.