Introduction to Trigonometry Test

Result

Introduction to Trigonometry Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

The value of tan 5° tan 25° tan 30° tan 65° tan 85° is

A
1/√2

B
1/√3

C
1/√7

D
1/√5

Solution

Grouping terms in such a way that the angles involved are complementary i.e., their sum is 90°, we have

= tan 5° tan 25° tan 30° tan 65° tan 85°

= (tan 5° tan 85°) (tan 25° tan 65°) tan 30°

= (tan 5° cot 5°) (tan 25° cot 25°) tan 30° [ tan 85° = tan (90° - 5°) = cot 5°, tan 65° = tan (90° - 25°) = cot 25°]

= 1 x 1 x 1/√3

= 1/√3
Question 2
1 / -0

tan 1° tan 2° tan 3° …….. tan 89° = ?

A
-1

B
1

C
0

D
None of these

Solution

We have,

= tan 1° tan 2° tan 3° …….. tan 89°

= tan (90° - 89°) tan (90° - 88°) tan (90° - 87°) …….. tan 87° tan 88° tan 89°

= cot 89° cot 88° cot 87° ……. tan 87° tan 88° tan 89°

= (cot 89° tan 89°) (cot 88° tan 88°) (cot 87° tan 87°) ……..(cot 44° tan 44°) tan 45°

= 1 x 1 x 1 ….. x 1 [cot θ tan θ = 1 and tan 45° = 1]

= 1
Question 3
1 / -0

If tan 2A = cot (A - 18°), where 2A is an acute angle then ∠A = ?

A
42°

B
40°

C
32°

D
36°

Solution

tan 2A = cot (A - 18°)

cot (90° - 2A) = cot (A - 18°) [ tan A = cot (90° - A)]

90° - 2A = A - 18°

90° + 18° = A + 2A

108° = 3A

3A = 108°

A = 108°/3

A = 36°
Question 4
1 / -0

Which of the following is correct some θ such that 0° ≤ θ < 90°

A
1/secθ > 1

B
1/secθ < 1

C
secθ = 0

D
1/cosθ < 1

Solution

1/secθ = cos θ and value of cos θ ranges from 0 to 1
Question 5
1 / -0

sin θ = 4/3 for some angle θ, is:

A
True

B
False

C
May be true for certain cases

D
None of these

Solution

sin θ ranges from 0 to 1. It can't be greater than 1.