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Introduction To Trigonometry Class 10 NCERT Solutions

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

Chapter 8 Introduction To Trigonometry Download in pdf

Download NCERT Solutions for Class 10 Mathematics

Chapter 8 Introduction To Trigonometry

(Link of Pdf file is given below at the end of the Questions List)

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


 


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