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Chapter 8 Introduction to Trigonometry Important Questions for CBSE Class 10 Maths Board Exams

This page contains Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry, which will help the students to prepare for the CBSE Class 10 maths Board exam 2022-23. It help the doing better in their maths paper. Extra questions of Chapter 8 Introduction to Trigonometry given here which are based on the pattern of CBSE NCERT book. Students will learn about the entire syllabus and learn how to solve problems in preparation for the exams.
Chapter 8 Introduction to Trigonometry Important Questions for CBSE Class 10 Maths Board Exams

Important Questions for Chapter 8 Introduction to Trigonometry Class 10 Maths


Important Questions Class 10 Maths contains all types of questions which could be asked in the examination. We have included short answer questions as well as long answer questions for Chapter 8 Introduction to Trigonometry. For more exercise, students can also refer to the Introduction to Trigonometry NCERT Solutions. Students can solve these questions and check their answers on the website. All the questions are solved in detail.

Introduction to Trigonometry Class 10 Maths Important Questions Very Short Answer (1 Mark)

1. If sec θ + tan θ = 7, then evaluate sec θ – tan θ.

Solution

We know that,
sec2θ – tan2θ = 1
⇒ (sec θ + tan θ) (sec θ – tan θ) = 1
⇒  (7) (sec θ – tan θ) = 1 …[sec θ + tan θ = 7]
∴ sec θ – tan θ = 1/7


2. Prove that

(1+tan A - sec A) (1+tan A + sec A) = 2 tan A

Solution

LHS = (1+tan A)2 - sec2 A

= 1+ tan2 A + 2 tan A - sec2 A

= sec2 A + 2 tan A - sec2 A

= 2 tan A = RHS


3. If tan A = cot B, then find the value of (A+B)

Solution

We have,

tan A = cot B

tan A = tan(90°-B)

A = 90° - B

Thus, A + B = 90°


4. If sinθ + sin2θ = 1 then prove that cos2θ + cos4θ = 1.

Solution

sinθ + sin2θ = 1

⇒ sinθ + (1-cos2θ) = 1

⇒ sinθ - cos2θ = 0

⇒ sinθ = cos2θ

Squaring both sides, we get

sin2θ = cos4θ

⇒ 1 - cos2θ = cos4θ

⇒ cos4θ + cos2θ = 1


5. If tan θ + cot θ = 5, find the value of tan2θ + cotθ.

Solution

tan θ + cot θ = 5 …[Given]
⇒ tan2θ + cot2θ + 2 tan θ cot θ = 25 …[Squaring both sides]
⇒ tan2θ + cot2θ + 2 = 25
∴ tan2θ + cot2θ = 23


6. If sec 2A = cosec (A – 27°) where 2A is an acute angle, find the measure of ∠A.

Solution

sec 2A = cosec (A – 27°)
⇒ cosec(90° – 2A) = cosec(A – 27°) …[∵ sec θ = cosec (90° – θ)]
⇒ 90° – 2A = A – 27°
⇒ 90° + 27° = 2A + A
⇒ 3A = 117°
∴ ∠A = 117°/3 = 39°


7. Evaluate: sin2 19° + sin271°.

Solution

sin219° + sin271°
= sin219° + sin2 (90° – 19°) …[∵ sin(90° – θ) = cos θ]
= sin2 19° + cos2 19° = 1 …[∵ sin2 θ + cos2 θ = 1]


8. In a triangle ABC, write cos(B+C/2) in terms of angle A.

Solution

In a triangle,
A+B+C = 180°
B+C = 180° - A

9. If secθ sinθ = 0, then find the value of θ.
Solution
We have,
secθ sinθ = 0
Thus θ = 0

10. Find the value of sin241° + sin249°

Solution

sin241° + sin249°

= sin2(90°-49°) + sin249°

= cos249° + sin249°

= 1


11. If tan A = cot B, prove that A + B = 90°.

Solution

tan A = cot B
∴ tan A = tan (90° − B)
⇒ A = 90° − B
⇒ A + B = 90°


12. Express sin 67° + cos 75° in terms of ratios of angles between 0° and 45°.

 Solution

∵ 67° = 90° − 23° and 75° = 90° − 15° 

∴ sin 67° + cos 75°

= sin (90° − 23°) + cos (90° − 15°)

= cos 23° + sin 15°


13. What is the value of sinθ. cos(90° - θ) + cosθ . sin(90° - θ)?

Solution

sinθ ·cos(90° − θ) + cosθ · sin(90° − θ)

= sinθ · sinθ + cosθ · cosθ [∵ cos(90° − θ) = sinθ , sin(90° − θ) = cos θ]
= sin2 θ + cos2 θ
= 1


14. If tan θ = cot (30° + θ ), find the value of θ .

Solution

We have,
tan θ = cot (30° + θ)
= tan [90° − (30° + θ)]
= tan [90° − 30° − θ]
= tan (60° − θ)
⇒ θ = 60° − θ
⇒ θ + θ = 60°
⇒ 2θ = 60°
⇒ θ = 60°/2
⇒ θ = 30°


15. If sin 3θ = cos (θ - 6)° and 3θ and (θ - 6)° are acute angles, find the value of θ.

Solution

We have,
sin3θ = cos(θ − 6)° = sin[90°−(θ − 6)°] ∵ [sin (90° − θ) = cos θ]
⇒ 3θ = 90° − (θ − 6)°
⇒ 3θ = 90° − θ + 6°
⇒ 3θ + θ = 96°
⇒ 4θ = 96°/4
⇒ θ = 24°


16. Show that: tan 10° tan 15° tan 75° tan 80° = 1

Solution

We have,

L.H.S. = tan 10° tan 15° tan 75° tan 80°
= tan (90° − 80°) tan 15° tan (90° − 15°) tan 80°
= cot 80° tan 15 cot 15° tan 80°
= (cot 80° × tan 80°) × (tan 15° × cot 15°)
= 1× 1
= 1 = R.H.S.


Introduction to Trigonometry Class 10 Maths Important Questions Short Answer-I (2 Marks)

17. If tan 2A = cot (A - 18°), where 2A is an acute angle, find the value of A.
Solution
tan 2A = cot (A - 18°)
⇒ cot (90° - 2A) = cot (A - 18°) [∵ cot (90° -0) - tan θ]
⇒ 90° - 2A = A - 18°
⇒ 3A = 108°
⇒ A = 108°/3
⇒ A = 36°

18. Prove that 1 + cot2α/1+cosecα = cosecα
Solution
= 1 + cosec α - 1
= cosec α

19. Prove that: sinA-2sin3A/2cos3A-cosA = tan A
Solution
= tan A

20. Show that tan4θ + tan2θ = sec4θ - sec2θ

Solution

tan4θ + tan2θ = tan2θ (1+ tan2θ)

= tan2θ × sec2θ

= (sec2θ - 1) sec2θ

= sec4θ - sec2θ


21. Find the value of (sin233° + sin257°)

Solution

sin233° + sin257°
⇒ sin233° + sin2( 90° - 33°)
⇒ sin233° + cos233 [Using sin(90° - θ) = cos θ]
⇒ 1 [Using sin2θ + cos2 θ = 1]


22. If ∆ABC is right angled at B, what is the value of sin (A + C).

Solution

∠B = 90° …[Given]
∠A + ∠B + ∠C = 180° …[Angle sum property of a ∆]
⇒ ∠A + ∠C + 90° = 180°
⇒ ∠A + ∠C = 90°
∴ sin (A + C) = sin 90° = 1 …(taking sin both side)


23. If tan θ = a/x, find the value of x/√a2+x2

Solution


24. If x = p sec θ + q tan θ and y = p tan θ + q sec θ, then prove that x2 – y2 = p2 – q2.

Solution

L.H.S. = x2 – y2
= (p sec θ + q tan θ)2 – (p tan θ + q sec θ)2
= p2 sec2θ + q2 tan2θ + 2 pq secθ tanθ - (p2 tan2θ + q2 sec2θ + 2pq secθ tanθ)
= p2 sec2θ + 2 tan2θ + 2pq secθ tanθ – p2 tan2θ – q2sec2θ – 2pq secθ tanθ
= p2(sec2θ – tan2θ) – q2(sec2θ – tan2θ)
= p2 – q2 …[sec2 θ – tan2 θ = 1]
= R.H.S.


25. If x = a cos θ – b sin θ and y = a sin θ + b cos θ, then prove that a2 + b2 = x2 + y2.

Solution

R.H.S. = x2 + y2
= (a cos θ – b sin θ)2 + (a sin θ + b cos θ)2
= a2cos2θ + b2 sin2θ – 2ab cosθ sinθ + a2 sin2θ + b2 cos2θ + 2ab sinθ cosθ
= a2(cos2 θ + sin2θ) + b2 (sin2 θ + cos2 θ)
= a2 + b2 = L.H.S. …[∵ cos2 θ + sin2 θ = 1]


26. Express cot 75° + cosec 75° in terms of trigonometric ratios of angles between 0° and 30°.

Solution

cot 75° + cosec 75°
= cot(90° – 15°) + cosec(90° – 15°)
= tan 15° + sec 15° …[cot(90°-A) = tan A]
= cosec(90° – A) = sec A


27. If cos (A + B) = 0 and sin (A – B) = 3, then find the value of A and B where A and B are acute angles.

Solution

Putting the value of B in (i), we get

⇒ A = 30° + 30° = 60°

∴ A = 60°, B = 30°


28. If sin (A + 2B) = √3/2 and cos (A + 4B) = 0, A > B, and A + 4B ≤ 90° then find A and B.

Solution

sin (A + 2B) = √3/2
⇒ sin(A + 2B) = sin 60°
Hence, A + 2B = 60° ...(i)
Also, we have
cos(A + 4B) = 0
⇒ cos(A + 4B) = cos 90°
⇒ A + 4B = 90° ...(ii)
Subtracting (ii) from (i), we have
-2B = - 30°
⇒ B = 15°
Put B = 15° in eq. (i), we have
A + 2(15°) = 60°
⇒ A + 30° = 60°
⇒ A = 30°


Introduction to Trigonometry Class 10 Maths Important Questions Short Answer-II (3 Marks)

29. Prove that:

tanθ/1-tanθ - cotθ/1-cotθ = cosθ+sinθ/cosθ-sinθ

Solution


30. Prove that: tan2A/tan2A-1 + cosec2A/sec2A-cosec2A = 1/1-2cos2A
Solution

31. Prove that: √1-cosA/√1+cosA = cosecA - cotA
Solution
= cosecA - cotA

32. If cos x = cos 40° . sin 50° + sin 40°. cos 50°, then find the value of x.

Solution

cos x = cos 40° sin 50° + sin 40° cos 50°
⇒ cos x = cos 40° sin(90° – 40°) + sin 40°.cos(90° – 40°)
⇒ cos x = cos2 40° + sin2 40°
⇒ cos x = 1 …[∵ cos2 A + sin2 A = 1]
⇒ cos x = cos 0°
⇒ x = 0°


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