# Trigonometry- MCQ Test 1

### 1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of tan C is:

AB = 24 cm and BC = 7 cm
tan C = perpendicular / base tan C = 24/7

### 2. (Sin 30°+cos 60°)-(sin 60° + cos 30°) is equal to:

sin 30° = ½, sin 60° = √3/2, cos 30° = √3/2 and cos 60° = ½
Putting these values, we get:
(½+½)-(√3/2+√3/2)
= 1 – [(2√3)/2]
= 1 – √3

1 – cos²A=sin²A

### 4. If cos X = ⅔ then tan X is equal to:

1 + tan²X = sec²X
And sec X = 1/cos X = 1/(⅔) = 3/2
Hence,
1 + tan²X = (3/2)² = 9/4
tan²X = (9/4) – 1 = 5/4
tan X = √5/2

### 5. sin 2A = 2 sin A is true when A =

Explanation/Answer:-sin 2A = 2 sin A
=> 2sinAcosA = 2 sin A
=> sinAcosA = sin A
=> cosA = 1
=> cosA = cos0°
=> A = 0°

### 6. The value of cos 0°. cos 1°. cos 2°. cos 3°… cos 89° cos 90° is

Explanation/Answer:- cos 0°. cos 1°. cos 2°. cos 3°… cos 89°. cos 90°
cos 90°=0

### 7. If x tan 45° sin 30° = cos 30° tan 30°, then x is equal to

Explanation/Answer:- x tan 45° sin 30° = cos 30° tan 30°
Put tan 45°=1, sin 30°=1/√2,cos 30°=√3/2 tan 30°=1/√3

### 8. If x and y are complementary angles, then

Explanation/Answer:- If x and y are complementary angles, then X=90-Y
sec x=sec(90-Y)= cosec y

### 9. If A and (2A – 45°) are acute angles such that sin A = cos (2A – 45°), then tan A is equal to

Explanation/Answer:- sin A = cos (2A – 45°)
cos(90°- A )= cos(2A – 45°)
(90°- A )= (2A – 45°)
3A = 135°
A = 45°
tan 45°=1