# 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 ChapterNCERT Exercise Solution Related Study NCERT Solution for Class 10 MathsNCERT Revision Notes for Class 10 MathsImportant Questions for Class 10 MathsMCQ for Class 10 MathsNCERT 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

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