# Mathematics Questions and Answers – Trigonometric Ratios of Specific Angles – 2

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This set of Mathematics Written Test Questions and Answers for Class 10 focuses on “Trigonometric Ratios of Specific Angles – 2”.

1. What is the value of tan θ when θ is 30°?
a) $$\frac {1}{\sqrt 3}$$
b) 1
c) 0
d) 2

Explanation: Tan θ = $$\frac {Sin \, \theta}{Cos \, \theta} = \frac {Sin \, 30^{\circ }}{Cos \, 30^{\circ }}$$
= $$\frac {1/2}{\sqrt 3/2}$$
= $$\frac {1}{2} \times \frac {2}{\sqrt 3}$$
= $$\frac {1}{\sqrt 3}$$

2. What is the value of sec 0° + cosec 90°?
a) 0
b) 2
c) 1
d) ∞

Explanation: The value of sec 0° is 1 and the value of cosec 90° is 1.
sec 0° + cosec 90° = 1 + 1
= 2

3. Evaluate sin2 30.
a) 2
b) 0
c) $$\frac {1}{4}$$
d) ∞

Explanation: Sin2 30° = ($$\frac {1}{2}$$)2
= $$\frac {1}{4}$$

4. What is the value of cos 45° + sin 45°?
a) 1
b) $$\frac {1}{4}$$
c) $$\frac {3}{4}$$
d) √2

Explanation: cos 45° + sin 45° = $$\frac {1}{\sqrt 2} + \frac {1}{\sqrt 2}$$
= 2 × $$\frac {1}{\sqrt 2}$$ ( Since, 2 = √2 × √2 )
= √2

5. 135° lies in the third quadrant.
a) True
b) False

Explanation: A plane is divided into four quadrants. The second quadrant ranges from 90° to 180°. So, 135° lies between 90° to 180° which is in the second quadrant of a plane not in the third quadrant.

6. What is the value of sec 60° – $$\frac {1}{cosec \, 90^{\circ }}$$?
a) Cot 30°
b) Cot 45°
c) Cot 60°
d) Tan 60°

Explanation: sec 60° – $$\frac {1}{cosec \, 90^{\circ }}$$ = 2 – $$\frac {1}{1}$$
= 1
= cot 45°

7. Tan 90° is _____
a) Not defined
b) 1
c) 0
d) 2

Explanation: Tan 90° = $$\frac {Sin \, 90^{\circ }}{Cos \, 90^{\circ }} = \frac {1}{0}$$
Any number divided with 0 is always not defined.

8. The value of sin2 90° + √2 cos 45° + √3 cot 30° is_____
a) 2
b) 0
c) 4
d) 5

Explanation: sin2 90° + √2 cos 45° + √3 cot 30° = 12 + √2 ($$\frac {1}{\sqrt 2}$$) + √3 (√3)
= 1 + 1 + 3
= 5

9. The product of sec 30° and cos 60° is _____
a) 0
b) 2
c) 1
d) $$\frac {1}{\sqrt 3}$$

Explanation: (sec 30°) (cos 60°) = $$( \frac {2}{\sqrt 3}) ( \frac {1}{2} )$$
= $$\frac {1}{\sqrt 3}$$

10. $$\frac {1 – cot \, 45^{\circ}}{1+ cot \, 45^{\circ}}$$ is _____
a) 0
b) 2
c) 3
d) 1

Explanation: $$\frac {1 – cot \, 45^{\circ}}{1 + cot \, 45^{\circ}} = \frac {1-1}{1+1}$$
= $$\frac {0}{1}$$
= 0

Sanfoundry Global Education & Learning Series – Mathematics – Class 10.

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