# Introduction To Trigonometry Class 10 NCERT Solutions

NCERT Solutions Class 10 Mathematics Chapter 8 Introduction To Trigonometry Download In Pdf

## In this pdf file you can see answers of following Questions

### EXERCISE 8.1

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

Question 2. In Fig. 8.13, find tan P – cot R.

Question 3. If sin A = 3 , 4 calculate cos A and tan A.

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

Question 5. Given sec θ = 13 , 12 calculate all other trigonometric ratios.

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

Question 7. If cot θ = 7 , 8 evaluate :
(i) (1 sin ) (1 sin ) , (1 cos ) (1 cos ) + θ − θ + θ − θ
(ii) cot2 θ

Question 8. If 3 cot A = 4, check whether 2 2 1 tan A 1 + tan A − = cos2 A – sin2A or not.

Question 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

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

Question 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.

### EXERCISE 8.2

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

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

Question 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

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

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

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

Question 4. If tan A = cot B, prove that A + B = 90°. 5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

#### EXERCISE 8.4

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

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

Question 3. Evaluate :
(i) 2 2 2 2 sin 63 sin 27 cos 17 cos 73 ° + ° ° + °
(ii) sin 25° cos 65° + cos 25° sin 65°

Question 4. Choose the correct option. Justify your choice.
(i) 9 sec2 A – 9 tan2 A =
(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) 2 2 1 tan A 1 + cot A + =
(A) sec2 A
(B) –1
(C) cot2 A
(D) tan2 A

Question 5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosec θ – cot θ)2 = 1 cos 1 cos − θ + θ
(ii) cos A 1 sin A 2 sec A
(iii) tan cot 1 sec cosec 1 cot 1 tan θ θ + = + θ θ − θ − θ [Hint : Write the expression in terms of sin θ and cos θ]
(iv) 1 sec A sin2 A sec A 1 – cos A + = [Hint : Simplify LHS and RHS separately]
(v) cos A – sin A + 1 cosec A + cot A, cos A + sin A – 1 = using the identity cosec2 A = 1 + cot2 A.
(vi) 1 sinA sec A + tan A 1 – sin A + =
(vii) 3 3 sin 2 sin tan 2 cos cos θ − θ = θ θ − θ
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
(ix) (cosec A – sin A)(sec A – cos A) 1 tanA + cot A = [Hint : Simplify LHS and RHS separately]
(x) 2 2 2 1 tan A 1 tanA 1 + cot A 1 – cot A = tan2 A ### NCERT Books Free Pdf Download for Class 5, 6, 7, 8, 9, 10 , 11, 12 Hindi and English Medium

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