NCERT Solution for Class 10 Mathematics Chapter 8 Trigonometry


Chapter Name

NCERT Solution for Class 10 Maths Chapter 8 Trigonometry

Topics Covered

  • Short Revision for the Chapter
  • NCERT Exercise Solution

Related Study

  • NCERT Solution for Class 10 Maths
  • NCERT Revision Notes for Class 10 Maths
  • Important Questions for Class 10 Maths
  • MCQ for Class 10 Maths
  • NCERT Exemplar Questions For Class 10 Maths

Short Revision for Trigonometry

  1. Trigonometry is the study of relationships between the sides and angles of a triangle. 

  2. sinθ = Perpendicular/Hypotenuse ,
    cosθ = Base/Hypotenuse,
    tanθ = Perpendicular/Base,
    cotθ = Base/Perpendicular,
    secθ = Hypotenuse/Base,
    cosecθ = Hypotenuse/Perpendicular. 
  3. cotθ = 1/tanθ,
    secθ = 1/cosθ,
    cosecθ = 1/sinθ
    tanθ = sinθ/cosθ
    cotθ = cosθ/sinθ
  4. sin(90° - θ) = cosθ,  cos(90° - θ) = sinθ,
    tan(90° - θ) = cotθ,  cot(90° - θ) = tanθ,
    sec(90° - θ) = cosec θ,  cosec(90° - θ) = secθ. 
  5. sin2θ + cos2θ = 1, sec2θ – tan2θ = 1, cosec2θ – cot2θ = 1.
  6. The value of sinθ or cosθ, is 0 or more but never exceeds 1. 
  7. The value of secθ or cosecθ is always greater than or equal to 1.

NCERT Exercise Solutions 

Exercise 8.1 

1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C


2. In Fig. 8.13, find tan P – cot R



3. If sin A = 3/4, Calculate cos A and tan A.

4. Given 15 cot A = 8, find sin A and sec A.

5. Given sec θ = 13/12 Calculate all other trigonometric ratios

6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Solution

∠A and ∠B are two acute angles either in same right - angled triangle or in different. 
Case I : Let ∠A and ∠B are acute angles in right △ABC. 

7. If cot θ = 7/8, evaluate :
(i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ)
(ii) cot2 θ

8. If 3 cot A = 4, check whether (1-tanA)/(1+tan2 A) = cos2 A – sin A or not.


9. In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C

10. In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P

Solution

PQ = 5cm and PR + QR = 25 cm 
Let RQ = x, then PR = 25 - x 
Using Pythagoras theorem, 

11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of angle A.
(iii)cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 4/3 for some angle θ.

Solution
(i) False, As tan A = p/b where perpendicular is not always less than base in a right - angled triangle. 
(ii) True, sec A = 12/5, true as cos A = 5/12 < 1, true. 
(iii) False, cos A is abbreviation used fro the cosine of angle A. 
(iv) False, cot A ≠ cot × A 
(v) False, as sin θ is always less than or equal to 1. 

Exercise 8.2 

1. Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60



2. Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan230° =
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) 1-tan245°/1+tan245° =
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin 2A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) 2tan30°/1-tan230° =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

3. If tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° < A + B ≤ 90°; A > B, find A and B.


4. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.


Exercise 8.3 

1. Evaluate :
(i) sin 18°/cos 72°
(ii) tan 26°/cot 64°
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°

2. Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0

3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

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

Solution

tan A = cot B = tan (90° - B)
⇒ A = 90° - B ⇒ A + B = 90°

Hence, proved. 


5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.


6. If A, B and C are interior angles of a triangle ABC, then show that sin (B+C/2) = cos A/2

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

Solution

Consider sin 67° + cos 75°
= sin(90° - 23°) + cos(90° - 15°) = cos 23° + sin 15° . 

Exercise 8.4 

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

2. Write all the other trigonometric ratios of ∠A in terms of sec A.


3. Evaluate:
(i) (sin263° + sin227°)/(cos217° + cos273°)
(ii) sin 25° cos 65° + cos 25° sin 65°

4. Choose the correct option. Justify your choice.
(i) 9 sec2A – 9 tan2A =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0 (B) 1 (C) 2 (D) – 1
(iii) (sec A + tan A) (1 – sin A) =
(A) sec A (B) sin A (C) cosec A (D) cos A
(iv) 1+tan2A/1+cot2A = 
(A) secA (B) -1 (C) cot2A (D) tan2A


5. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.
(i) (cosec θ – cot θ)= (1-cos θ)/(1+cos θ)


(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A


(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]


(iv) (1 + sec A)/sec A = sin2A/(1-cos A)
[Hint : Simplify LHS and RHS separately]

(v) ( cos A–sin A+1)/( cos A +sin A–1) = cosec A + cot A, using the identity cosec2A = 1+cot2A.


(vi) √(1+ sinA/1-sinA) = sec A + tan A


(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ


(viii) (sin A + cosec A)+ (cos A + sec A)2 = 7+tan2A+cot2A


(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]


(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 =tan2A

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