NCERT Solutions for Class 10 Maths Exercise 8.3 (NEW SESSION)

Solve the followings Questions.

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

chapter 8-Introduction to Trigonometry Exercise 8.4/image004.png
chapter 8-Introduction to Trigonometry Exercise 8.4/image005.png

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

Answer:

chapter 8-Introduction to Trigonometry Exercise 8.4/image021.png
chapter 8-Introduction to Trigonometry Exercise 8.4/image006.png

3. Evaluate :

(i) (sin263° + sin227°)/(cos217° + cos273°)
(ii)  sin 25° cos 65° + cos 25° sin 65°

Answer:

 (i) (sin263° + sin227°)/(cos217° + cos273°)

Maths sample paper for class 10 /image039.png

(ii)  sin 25° cos 65° + cos 25° sin 65°

Maths sample paper for class 10 /image041.png

Choose the correct option. Justify your choice.

4. (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) (secA + tanA) (1 – sinA) =
(A) secA           (B) sinA   (C) cosecA      (D) cosA
(iv) 1+tan2A/1+cot2A =
(A) sec2A          (B) -1      (C) cot2A                (D) tan2A

Answer:

(i)  (i) 9 sec2A – 9 tan2A =

(A) 1                 (B) 9        (C) 8                (D) 0 

Maths sample paper for class 10 /image058.png

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0                 (B) 1        (C) 2                (D) – 1

Maths sample paper for class 10 /image048.png

(iii) (secA + tanA) (1 – sinA) =
(A) secA           (B) sinA   (C) cosecA      (D) cosA

Maths sample paper for class 10 /image061.png

(iv) 1+tan2A/1+cot2A =
(A) sec2A          (B) -1      (C) cot2A                (D) tan2A

Maths sample paper for class 10 /image063.png

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 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 + sin A/1 – sin A = sec A+ tan A

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

(viii) (sin A + cosec A)2 + (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

Answer:

(i) (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)

NCERT solutions /image089.png

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

NCERT solutions /image091.png

(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]

NCERT solutions /image093.png
NCERT solutions /image095.png

(iv) (1 + sec A)/sec A = sin2A/(1-cos A)  

[Hint : Simplify LHS and RHS separately] 

NCERT solutions /image100.png
NCERT solutions /image101.png

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

NCERT solutions /image121.png
NCERT solutions /image122.png

(vi) √1 + sin A/1 – sin A = sec A+ tan A

LHS = 1 + sin A/(1 – sin A) …..(1)

Multiplying and dividing by (1 + sin A)

⇒ (1 + sin A)(1 + sin A/1 – sin A)(1 + sin A)

= (1 + sin A)²/(1 – sin² A) [a² – b² = (a – b)(a + b)]

= (1 + sinA)/1 – sin² A

= 1 + sin A/cos² A

= 1 + sin A/cos A

= 1/cos A + sin A/cos A

= sec A + tan A

= R.H.S

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

NCERT solutions /image141.png

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

NCERT solutions /image147.png

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

 [Hint : Simplify LHS and RHS separately]

NCERT solutions /image156.png
NCERT solutions /image157.png

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

NCERT solutions /image166.png
NCERT solutions /image167.png
CHAPTER NAMEOLD NCERTNEW NCERT 
Real NumbersEXERCISE 1.1 
EXERCISE 1.21.1CLICK HERE
EXERCISE 1.31.2CLICK HERE
EXERCISE 1.4
PolynomialsEXERCISE 2.12.1CLICK HERE
EXERCISE 2.22.2CLICK HERE
EXERCISE 2.3
EXERCISE 2.4
Pair of Linear Equations in Two VariablesEXERCISE 3.1
EXERCISE 3.23.1CLICK HERE
EXERCISE 3.33.2CLICK HERE
EXERCISE 3.43.3CLICK HERE
EXERCISE 3.5
EXERCISE 3.6
EXERCISE 3.7
Quadratic EquationsEXERCISE 4.14.1CLICK HERE
EXERCISE 4.24.2CLICK HERE
EXERCISE 4.3
EXERCISE 4.44.3CLICK HERE
Arithmetic ProgressionsEXERCISE 5.15.1CLICK HERE
EXERCISE 5.25.2CLICK HERE
EXERCISE 5.35.3CLICK HERE
EXERCISE 5.45.4 (Optional)CLICK HERE
TrianglesEXERCISE 6.16.1CLICK HERE
EXERCISE 6.26.2CLICK HERE
EXERCISE 6.36.3CLICK HERE
EXERCISE 6.4
EXERCISE 6.5
EXERCISE 6.6
Coordinate GeometryEXERCISE 7.17.1CLICK HERE
EXERCISE 7.27.2CLICK HERE
EXERCISE 7.3
EXERCISE 7.4
Introduction to TrigonometryEXERCISE 8.18.1CLICK HERE
EXERCISE 8.28.2CLICK HERE
EXERCISE 8.3
EXERCISE 8.48.3CLICK HERE
Some Applications of TrigonometryEXERCISE 9.19.1CLICK HERE
CirclesEXERCISE 10.110.1CLICK HERE
EXERCISE 10.210.2CLICK HERE
Construction
Areas Related to CirclesEXERCISE 12.1
EXERCISE 12.211.1CLICK HERE
EXERCISE 12.3
Surface Areas and VolumesEXERCISE 13.112.1CLICK HERE
EXERCISE 13.212.2CLICK HERE
EXERCISE 13.3
EXERCISE 13.4
EXERCISE 13.5
StatisticsEXERCISE 14.113.1CLICK HERE
EXERCISE 14.213.2CLICK HERE
EXERCISE 14.313.3CLICK HERE
EXERCISE 14.4
ProbabilityEXERCISE 15.114.1CLICK HERE
EXERCISE 15.2

Leave a Reply

Your email address will not be published. Required fields are marked *

error: Content is protected !!