Trigonometry Questions For All Exams With Complete Solution

Today, in this article you will learn how to solve specific types of questions related to trigonometry. As far as I know, these types of questions are asked in every exam whether it’s competitive or board exam.

Moreover, trigonometry is the most important topic in Mathematics.

Without this chapter Mathematics seems incomplete. Because, we learn solving different types of questions in this chapter.

Whatever the exam is, trigonometry questions are definitely asked. It means, the weightage of asking trigonometry questions is good enough in any exam.

If you are going to face board or competitive exams, definitely this article will prove to be the best and worthy for you.

let’s start the topic.

Firstly, I’ll tell you some of the important formulas, which really help to solve trigonometry questions.

If you don’t learn or focus on these basic formulas, definitely you won’t be able to solve trigonometry questions.

Here are the most important and basic formulas of trigonometry.

  • sin²A+cos²A = 1
  • 1-sin²A = cos²A
  • 1-cos²A = sin²A
  • 1+tan²A = Sec²A
  • 1+cot²A = cosec²A
  • sec²A-tan²A = 1
  • cosec²A- cot²A = 1
  • sin2A = 2sinAcosA
  • tan2A = 2tanA/1-tan²A
  • sin2A = 2tanA/1+tan²A
  • cos2A = cos²A-sin²A = 2cos²A-1 = 1-2sin²A = 1-tan²A/1+tan²A
  • sin3A = 3sinA-4sin³A
  • cos3A = 4cos³A-3cosA
  • tan3A = 3tanA-tan³A/1-3tan²A

All these are the basic formulas related to trigonometry. You must learn these to go ahead rightly.

Trigonometry Questions For All Exams With Complete Solution –

Now, we will focus on the most important types of the questions which are definitely asked in the examination.

Q1. What is the value of cos( 11π/6 ) ? –

Solution – cos ( 11π/6 )

\because π = 180°

= cos ( 11×180°/6 )

= cos ( 330° )

= cos ( 360-30 )

= cos 30°

= √3/2 ( Ans )

In trigonometry, this type of question is very common. You can assume other values instead of ( 11π/6 ) for practicing broadly.

If you are a student of maths and want to know about BSc, read this article by clicking the provided link.

Q2. What is the value of \frac{2( 1-sin^{2}A )cosec^{2}A}{cot^{2}A( 1+tan^{2}A )} – 1 ? –

Solution – \frac{2( 1-sin^{2}A )cosec^{2}A}{cot^{2}A( 1+tan^{2}A )} – 1

= \frac{2cos^{2}Acosec^{2}A}{cot^{2}Asec^{2}A} – 1

= \frac{2cos^{2}Asin^{2}A}{sin^{2}Acos^{2}Asec^{2}A} – 1

= 2cos²A-1

= cos2A ( Ans )

Q3. What is the value of sin²30°.cos²45°+2tan²30°-sec²60° ?-

Solution – sin²30°.cos²45°+2tan²30°-sec²60°

= (1/2)².(1/√2)²+2(1/√3)²-(2)²

= (1/4).(1/2)+2.(1/3)-4

= (1/8)+(2/3)-4

= (3+16-96)/24

= -77/24 ( Ans )

Q4. if sin A = 5/12 & tan A = 12/5, find the value of sin²A-tan²A.

Solution – sin²A-tan²A

= (5/12)²-(12/5)²

= (25/144)-(144/25)

= {(25)²-(144)²}/(144×25)

\because a²-b² = ( a + b ) ( a – b )

= {(25+144) (25-144)}/(144×25)

= (169×-119)/(144×25)

= 20111/3600 ( Ans )

Q5. 7sin²A + 3cos²A = 4, find the value of A.

Solution – 7sin²A + 3cos²A = 4

\Rightarrow 7sin²A + 3(1-sin²A) = 4

\Rightarrow 7sin²A + 3 – 3sin²A = 4

\Rightarrow 4sin²A + 3= 4

\Rightarrow 4sin²A = 1

\Rightarrow sin²A = 1/4

\Rightarrow sinA = 1/2

\Rightarrow sinA = sin30°

\Rightarrow A = 30° ( Ans )

Q6. If tan4A = cot(2A+30°), find the value of A.

Solution – tan4A = cot(2A+30°)

\Rightarrow tan4A = tan( 90°-2A-30°)

\Rightarrow tan4A = tan( 60°-2A )

\Rightarrow 4A = 60°-2A

\Rightarrow 6A = 60°

\Rightarrow A = 10° ( Ans )

Q7. If cos A + sin A = √2cosA, find out the value of ( cos A – sin A ).

Solution – cos A + sin A = √2cosA

squaring both sides

( cos A + sin A )² = ( √2cosA )²

\Rightarrow cos²A + sin²A + 2sinAcosA = 2cos²A

\Rightarrow 2sinAcosA = 2cos²A – cos²A – sin²A

\Rightarrow 2sinAcosA = cos²A – sin²A

\Rightarrow 2sinAcosA = ( cos A + sin A ) ( cos A – sin A )

\because cos A + sin A = √2cosA ( given )

\Rightarrow 2sinAcosA = √2cosA ( cos A – sin A )

\Rightarrow \frac{2sinAcosA}{\sqrt{2}cosA} = cos A – sin A

\Rightarrow cos A – sin A = \frac{2sinAcosA}{\sqrt{2}cosA} ( Ans )

Q8. What is the value of sec²A\frac{sin^{2}A-2sin^{4}A}{2cos^{4}A-cos^{2}A} ? –

Solution – sec²A – \frac{sin^{2}A-2sin^{4}A}{2cos^{4}A-cos^{2}A}

= sec²A – \frac{sin^{2}A(1-2sin^{2}A ) }{cos^{2}A(2cos^{2}A-1)}

\because 1-2sin²A = 2cos²A-1 = cos2A

= sec²A – \frac{sin^{2}A(cos2A ) }{cos^{2}A(cos2A)}

= sec²A – \frac{sin^{2}A}{cos^{2}A}

= sec²A – tan²A

= 1

Q9. Find the value of \frac{sinA}{1+cosA}+\frac{1+cosA}{sinA} ? –

Solution \frac{sinA}{1+cosA}+\frac{1+cosA}{sinA}

= \frac{sinA(1-cosA)}{(1+cosA)(1-cosA)}+\frac{(1+cosA)}{sinA}

= \frac{sinA(1-cosA)}{(1-cos^{2}A)}+\frac{(1+cosA)}{sinA}

= \frac{sinA(1-cosA)}{sin^{2}A}+\frac{(1+cosA)}{sinA}

= \frac{(1-cosA)}{sinA}+\frac{(1+cosA)}{sinA}

= \frac{(1-cosA+1+cosA)}{sinA}

= \frac{2}{sinA}

= 2cosecA ( Ans )

Q10. Find the value of \sqrt{1+cot^{2}A}.\sqrt{1-sin^{2}A}.\sqrt{sec^{2}A-1} ? –

Solution \sqrt{1+cot^{2}A}.\sqrt{1-sin^{2}A}.\sqrt{sec^{2}A-1}

= \sqrt{cosec^{2}A}.\sqrt{cos^{2}A}.\sqrt{tan^{2}A}

= cosec A. cos A. tan A

= \frac{1}{sinA}.\frac{cosA}{1}.\frac{sinA}{cosA}

= 1 ( Ans )

Q11. What is the value of 27° into Radian ? –

Solution – 27°

\because 1° = π/180° ( radian )

= 27°×π/180° radian

= 3π/20 radian ( Ans )

Q12. Find out the value of \frac{sin300.sec420}{tan225.sin210}.

Solution \frac{sin300.sec420}{tan225.sin210}

= \frac{sin(300-60).sec(360+60)}{tan(180+45).sin(180+30)}

= \frac{-sin60.sec60}{tan45.-sin30}

= \frac{-sin60.sec60}{tan45.-sin30}

= \frac{-\sqrt{3}/2\times 2}{1\times-1/2}

= 2√3 ( Ans )

Q13. \frac{sin39^{\circ}}{cos51^{\circ}}+2tan11°.tan31°.tan45°.tan59°.tan79-3(sin²21°+sin²69°) ? –

Solution \frac{sin39^{\circ}}{cos51^{\circ}}+2tan11°.tan31°.tan45°.tan59°.tan79-3(sin²21°+sin²69°)

= \frac{sin39^{\circ}}{sin( 90-51)^{\circ}}+2tan11°.tan31°.tan45°.cot( 90°-59° ).cot( 90°-79° )-3(sin²21+cos²(90°-69°)

= \frac{sin39^{\circ}}{sin39^{\circ}}+2tan11°.tan31°.tan45°.cot31°.cot11°-3(sin²21+cos²21)

= 1+2tan11°.tan31°.1/tan31°.1/tan11°-3×1

= 1+2×1×1-3

= 1+2-3

= 3-3

= 0

Q14. Prove that ( sec A – cos A )²+ ( cosec A – sin A )²- ( cot A – tan A )² = 1 ? –

Solution – L.H.S – ( sec A – cos A )²+ ( cosec A – sin A )²- ( cot A – tan A )²

= sec²A+cos²A-2secA.cosA+cosec²A+sin²A-2cosecA.sinA-( cot²A+tan²A-2cotAtanA )

= sec²A+cos²A-2secA.cosA+cosec²A+sin²A-2cosecA.sinA-cot²-tan²A+2cotA.tanA

= sec²A+cosec²A+cos²A+sin²A-2×1/cosA×cosA-2×1/sinA×sinA-cot²A-tan²A+2×1/tanA×tanA

= sec²A+cosec²A+1-2-2-cot²A-tan²A+2

= 1+tan²A+1+cot²A+1-2-2-cot²A-tan²A+2

= 1+1+1-2-2+2

= 1 ( R.H.S ) Proved.

Q15. If tan A – cot A = 0, then find the value of ( sin A – cos A ).

Solution – tan A – cot A = 0 ( given )

\Rightarrow tan A = cot A

\Rightarrow tan A = 1/tan A

\Rightarrow tan²A = 1

\Rightarrow tan A = 1

\Rightarrow tan A = tan 45°

\Rightarrow A = 45°

now, the value of ( sin A-cos A ) is

= sin 45°-cos45°

= 1/√2 – 1/√2

= 0 ( Ans )

20 Extra trigonometry questions for practice –

1. sin ( -330 ) – Ans is 1/2

2. cot ( 4π/3 ) – Ans is 1/√3

3. tan ( 7π/4 ) – Ans is -1

4. cot ( -5π/4 ) – Ans is -1

5. sin ( 5π/3 ) – Ans is -√3/2

6. {( sinA-2sin³A )/( 2cos³A-cosA )}²+1 – Ans is ( sec²A )

7. If {( cos²A )/( cot²A-cos²A )}= 3, then find the value of cot A+ cosec A – Ans is ( √3 )

8. If {( tanA-secA+1 )/( tanA+secA-1 )}secA = 1/k, then find the value of K – Ans is ( 1+sinA )

9. {( 1+cosA )²+sin²A}/{( cosec²A-1 )sin²A}- Ans is 2secA( 1+secA )

10. If 3sinA = 2cosA, then find the value of ( 4sinA-cosA )/( 4cosA+sinA ) – Ans is ( 5/14 )

11. ( sin30°-cos60°+cot²45° )/( cos30°-tan45°+sin90° ), find the value – Ans is ( 2√3/3 ) or ( 2/√3 )

12. 4sin²30°+3cot²60° – Ans is ( 2 )

13. tan6°tan36°tan84°tan54°tan45° – Ans is ( 1 )

14. { sec60°/( sec60°-1 )}+{ sec60°/( sec60°+1 ) } – Ans is ( 8/3 )

15. tan²60°+cosec²60°+sin²60° – Ans is ( 61/12 )

16. If sin A = 12/35, find the value of tan A, cos A & cot A.

17. In a triangle, right angled at B and cot A = 6/8, find the value of sin C & tan A.

18. In a right angled triangle, if AB = 9, BC = 12 then find the value of CA.

19. sin135°.cos225°.sin1125°/tan135°.sin315° – Ans is -1/2

20. cos135°.sin210°.cos150°/sin270°.cos315° – Ans is √3/4

Eventually, I hope that you must have got valuable questions from this article.

Well, I have taken all important types of questions related to trigonometry, nevertheless you must read books for making trigonometry perfect.

When you read trigonometry chapter from book, you get verity of trigonometry questions which really help you for exam purpose.

At last, if you really liked the article, please share it to those students who need it.

Thank you.

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