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Trigonometry questions are a major type of questions asked in competitive exams. These questions carry a weightage of 3-4 questions (6-8 marks) in SSC exams. To get a good rank in competitive exams, you should have a great practice of a variety of trigonometry questions

Here are some tips for solving Trigonometry questions: Clear the basic concepts, Memorise the key ratios, Memorise the identities and formulas, practice a variety of questions, and crack the question you should have a great understanding and practice.

So, we have attached 10 questions of Trigonometry for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Trigonometry</h2>

You can also download the Trigonometry questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

 Q:1 If cosec θ + cot θ = 2, then find the value of cosec θ.

1. (5/7)

2. (5/3)

3. (5/4)

4. (5/9)

(Difficulty: 3, Estimated Time: 20 Seconds) This was an easy one! Did you get it right?

Q:2 If sin(A + B) = 1 and cos(A – B) = (√3/2), where 90o ≥ A > B > 0o then find the value of tan ‘A’.

1. 0

2. 6

3. √3

4. √7

(Difficulty: 3, Estimated Time: 20 Seconds) In these questions, keep in mind the range of angle given in the question!

Q:3 If cosec A = (25/7), then find the value of sec A.

1. (27/13)

2. (25/24)

3. (29/17)

4. (29/17)

(Difficulty: 2, Estimated Time: 10 Seconds) Do you remember some Pythagorean triplets?

 Q:4 Find the value of:

\(\sqrt{\frac{1+\cos A}{1-\cos A}} \times \sqrt{1-\cos ^2 A}\)

1. 1 - sin A

2. 1 - cos A

3. 1 + cos A

4. 1 + sin A

(Difficulty: 3, Estimated Time: 20 Seconds) Should we increase the level?

Q:5 If sin θ + cos θ = √5 sin(90o – θ), then find the value of cot θ

1. 1

2. {(√5 – 1)/4}

3. (√3 + 4)

4. {(√5 + 1)/4}

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 If cotx – tanx = 3, then find the value of (cotx + tanx).

1. 12

2. √13

3. 7

4. √11

(Difficulty: 4, Estimated Time: 30 Seconds) This was a test of your concepts!

Q:7 If tan2θ = 1 – a2, then find the value of secθ + tan3θcosecθ

1. (2 + a2)(3/2)

2. (1 – a2)

3. (2 - a2)(3/2)

4. (1 + a2)

(Difficulty: 3, Estimated Time: 20 Seconds) This was a hard nut to crack, be prepared for such questions in exam!

Q:8 Find the value of cot 24o cot 36o cot 48o cot 60o cot 42o cot 54o cot 66o

1. 1/√2

2. 1/√3

3. 1/√7

4. 1/√5

(Difficulty: 2, Estimated Time: 15 Seconds) Another easy one!

Q:9 If cosec A – cot A = (1/5), then find the value of tan A.

1. 5/12

2. 2/7

3. 3/19

4. 13/23

(Difficulty: 3, Estimated Time: 20 Seconds) Yes, You know how to solve such questions!

Q:10 If sin A = (4/5) and sin B = (12/13), then find the value of sin (A – B).

1. –(9/69)

2. (11/112)

3. –(16/65)

4. (14/63)

(Difficulty: 2, Estimated Time: 15 Seconds) Do you remember the formulas? 

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (3)</td>
			<td>2. (3)</td>
			<td>3. (2)</td>
			<td>4. (3)</td>
			<td>5. (4)</td>
		</tr>
		<tr>
			<td>6. (2)</td>
			<td>7. (3)</td>
			<td>8. (2)</td>
			<td>9. (1)</td>
			<td>10. (3)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 3 i.e. 5/4

Given = cosec θ + cot θ = 2

We know, cosec2 θ – cot2θ = 1 …………. (I)

Or, (cosec θ + cot θ) × (cosec θ – cot θ) = 1

Or, cosec θ – cot θ = (1/2) …………… (II)

On adding equations (I) and (II), we have;

2 cosec θ = 2 + (1/2)

So, cosec θ = (5/4)

 

Q:2 The correct answer is option 3 i.e. √3

sin(A + B) = sin 90o

So, A + B = 90o …………… (I)

Also, cos(A – B) = cos 30o

So, A – B = 30o ……………. (II)

On adding equation (I) and (II), we have;

2A = 120o

So, A = 60o

So, tanA = tan60o = √3

 

Q:3 The correct answer is option 2 i.e. 25/24

cosecθ = (Hypotenuse/perpendicular)

secθ = (Hypotenuse/base)

Base2 = Hypotenuse2 – Percendicular2

Base2 = 252 – 72

Base2 = √576 = ±24 (Since, the length cannot be negative therefore, we will take the positive root only)

So, secA = (hypotenuse/base) = (25/24)

 

Q:4 The correct answer is option 3 i.e. 1 + cos A

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/CA_22.08.2023_D16.PNG" />

 

Q:5 The correct answer is option 4 i.e. {(√5 + 1)/4}

Given,

\( \begin{aligned} & \sin \theta+\cos \theta=\sqrt{5} \sin (90-\theta), \\ & =\sin \theta+\cos \theta=\sqrt{5} \cos \theta \\ & \text { Or, } \sin \theta=(\sqrt{5}-1) \cos \theta \\ & \text { Or, } \tan \theta=(\sqrt{5}-1) \\ & \cot \theta=\frac{1}{\sqrt{5}-1} \\ & \cot \theta=\frac{1}{\sqrt{5}-1} \times \frac{\sqrt{5}+1}{\sqrt{5}+1} \\ & \cot \theta=\frac{\sqrt{5}+1}{5-1} \\ & \cot \theta=\frac{\sqrt{5}+1}{4} \end{aligned} \)

 

Q:6 The correct answer is option 2 i.e. √13

Given,

\( \cot x-\tan x=3\) Let \(\cot x=y\) Then, \(\tan x=(1 / y) \)

Given, \( y-\frac{1}{y}=3 \)

Squaring on both sides,

\( \begin{aligned} & \left(y-\frac{1}{y}\right)^2=3^2 \\ & \text { Or, } y^2+\frac{1}{y^2}-2=9 \\ & y^2+\frac{1}{y^2}+2=9+4=13 \\ & \left(y+\frac{1}{y}\right)=\sqrt{13} \end{aligned} \)

tanx + cotx = √13

 

Q:7 The correct answer is option 3 i.e. (2 - a2)(3/2)

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/CA_22.08.2023_D17.PNG" />

We have, 

\( tan ^2 \theta=1-a^2\) So, \(1+\tan ^2 \theta=1-a^2+1\) Or, \(\sec ^2 \theta=\left(2-a^2\right)\) So, \(\sec \theta+\tan ^3 \theta \operatorname{cosec} \theta=\sec ^3 \theta=\left(2-a^2\right)^{\frac{3}{2}}\)

 

Q:8 The correct answer is option 2 i.e. 1/√3

cot 24o cot 36o cot 48o cot 60o cot 42o cot 54o cot 66o

= (cot 24o × cot 66o) × (cot 36o × cot 54o) × (cot 48o × cot 42o) × cot 60o

= 1 × 1 × 1 × (1/√3) …….. Since cot θ × cot (90o – θ) = 1

= (1/√3)

 

Q:9 The correct answer is option 1 i.e. 5/12

Given, cosec A – cot A = (1/5) – [I]

We know that, cosec2 A – cot2 A = 1

So, cosec A + cot A = (cosec2 A – cot2 A) ÷ (cosec A – cot A)

Or, cosec A + cot A = 1 ÷ (1/5) = 5 – [II]

Subtracting [I] from [II],

We have, cosec A + cot A – (cosec A – cot A) = 5 – (1/5)

Or, 2cot A = (24/5)

Or, cot A = (24/5) ÷ 2 = (12/5)

So, tan A = (5/12)

 

Q:10 The correct answer is option 3 i.e. –(16/65)

Given, sin A = (4/5)

So, perpendicular = 4 and hypotenuse = 5

So, base = √(52 – 42) = 3

So, cos A = (3/5)

Given, sin B = (12/13)

So, perpendicular = 12 and hypotenuse = 13

So, base = √(132 – 122) = 5

So, cos B = (5/13)

Sin (A – B) = sinAcosB – cosAsinB

= (4/5) × (5/13) – (3/5) × (12/13)

= (20/65) – (36/65)

= –(16/65)

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Trigonometry.

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