# Trigonometry MCQ Mock Test Online Free

Trigonometry MCQs are different types of questions that come in all types of exams. Trigonometry plays an important role in Maths section. If you want to clear your concepts, then you can practice Mathematics introduction online from our website www.sarkarijobalert.com Trigonometry MCQ Test Free Mochak Test. This Introduction to Trigonometry Class 10 MCQs questions examines the understanding and concept of the chapter. You can test your knowledge and evaluate yourself. We suggest that you give as many online free MCQ mock tests as possible so that your concepts can be cleared.

**Trigonometry MCQ Online Free Mock Test 1**

[ays_quiz id=’19’]

## Example **Trigonometry MCQ**

**1) Which of the following is the correct value of cot 10 ^{0}.cot 20^{0}.cot 60^{0}.cot 70^{0}.cot 80^{0}?**

- 1/√3
- √3
- -1
- 1

**Answer:** (a) 1/√3

**Explanation:** Here, we can apply the formula –

cot A. cot B = 1 (when A + B = 90^{0})

= (cot 20^{0} . cot 70^{0}) x (cot 10^{0} . cot 80^{0}) x cot 60^{0}

= 1 x 1 x 1/√3

= 1/√3

So, the correct value of cot 10^{0}.cot 20^{0}.cot 60^{0}.cot 70^{0}.cot 80^{0} = 1/√3

**2) If tan θ + cot θ = 2, then what is the value of tan ^{100} θ + cot^{100} θ?**

- 1
- 3
- 2
- None of the above

**Answer:** (c) 2

**Explanation:** Given tan θ + cot θ = 2

Put θ = 45^{0}, above equation will satisfy as,

1 + 1 = 2

So, θ = 45^{0},

= tan^{100} 45^{0} + cot^{100} 45^{0}

= 1^{100} + 1^{100}

= 2

**3) If x and y are complementary angles, then**

(a) sin x = sin y

(b) tan x = tan y

(c) cos x = cos y

(d) sec x = cosec y

**Answer**: d

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

A. 0

B. 1/√3

C. 1

D. √3

**Answer**: C

**5) If θ is said to be an acute angle, and 7 sin2 θ + 3 cos2 θ = 4, then what is the value of tan θ?**

- 1
- √3
- 1/√3
- None of the above

**Answer:** (c) 1/√3

**Explanation:** Given 7 sin^{2} θ + 3 cos^{2} θ = 4

=> 7 sin^{2} θ + 3 (1 – sin^{2} θ) = 4

=> 7 sin^{2} θ + 3 – 3sin^{2} θ = 4

Then, 4sin^{2} θ = 1

Or, sin θ = 1/2

So, θ = 30^{0}

Now, put θ = 30^{0} in tan θ, we will get,

tan θ = 1/√3

**6) Suppose cos θ + sin θ = √2 cos θ, then which of the following is the correct value of cos θ – sin θ?**

- √2 cos θ
- √2 sin θ
- -√2 cos θ
- -√2 sin θ

**Answer:** (b) √2 sin θ

**Explanation:** It is given that, cos θ + sin θ = √2 cos θ …..(i)

On squaring both sides, we will get,

(cos θ + sin θ)^{2} = (√2 cos θ)^{2}

=> cos^{2} θ + sin^{2} θ + 2 sin θ cos θ = 2 cos^{2} θ

Or, 2cos^{2} θ – cos^{2} θ – sin^{2} θ = 2 sinθ cosθ

=> cos^{2} θ – sin^{2} θ = 2 sin θ cos θ

=> (cos θ + sin θ) (cos θ – sin θ) = 2 sin θ cos θ

=> (√2 cos θ) (cos θ – sin θ) = 2 sin θ cos θ [from equation (i)]

=> (cos θ – sin θ) = 2 sinθ cosθ / √2 cos θ

= √2 sin θ

**7) The value of cos 180° is**

(a) 0

(b) 1

(c) -1

(d) infinite

**Answer**

Answer: (c) -1

Hint:

180 is a standard degree generally we all know their values but if we want to go theoretically then

cos(90 + x) = – sin(x)

So, cos 180 = cos(90 + 90)

= -sin 90

= -1 {sin 90 = 1}

So, cos 180 = -1

**8) In given figure, the length of AP is**

**Answer/ Explanation**

Answer: b

Explaination:

**9) If the value of tanP + secP = a, then what is the value of cosP?**

- 2a/a
^{2}+ 1 - a
^{2}+ 1/ 2a - a
^{2}– 1/ 2a - None of the above

**Answer:** **(a)** 2a/a^{2} + 1

**Explanation:** It is given that, tanP + secP = a ……(i)

As we know, the trigonometric identity, sec^{2} P – tan^{2} P = 1 {we assume θ = P}

So, we can apply the formula a^{2} – b^{2} = (a – b) (a + b)

=> (sec P – tan P) (sec P + tan P) = 1

=> (sec P – tan P) x a = 1

=> sec P – tan P = 1/a …..(ii)

So, from equation (i) and (ii), we will get –

2sec P = a + 1/a

sec P = a^{2}+1 / 2a

So, cos P = 2a / a^{2}+1 [as sec P = 1/cosP]

**10) Which of the following is the correct value of (3 / 1+tan ^{2} θ) + 2 sin^{2} θ + (1 / 1+cot^{2} θ)?**

- 3
- 9
- 6
- None of the above

**Answer:** (a) 3

**Explanation:** (3 / 1+tan^{2} θ) + 2 sin^{2} θ + (1 / 1+cot^{2} θ) = ?

According to the trigonometric identities, the given equation can be written as –

= 3/sec^{2} θ + 2 sin^{2} θ + 1/cosec^{2} θ

= 3cos^{2} θ + 2 sin^{2} θ + sin^{2} θ

= 3cos^{2} θ + 3sin^{2} θ

= 3(cos^{2} θ + sin^{2} θ)

= 3