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NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry (Exercise 8.3)

Q1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:
To convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas
We know that,
cosec2A – cot2A = 1
cosec2A = 1 + cot2A
Since cosec function is the inverse of sin function, it is written as
1/sin2A = 1 + cot2A
Now, rearrange the terms, it becomes
sin2A = 1/(1+cot2A)
Now, take square roots on both sides, we get
sin A = ±1/(√(1+cot2A)
The above equation defines the sin function in terms of cot function
Now, to express sec function in terms of cot function, use this formula
sin2A = 1/ (1+cot2A)
Now, represent the sin function as cos function
1 – cos2A = 1/ (1+cot2A)
Rearrange the terms,
cos2A = 1 – 1/(1+cot2A)
⇒cos2A = (1-1+cot2A)/(1+cot2A)
Since sec function is the inverse of cos function,
⇒ 1/sec2A = cot2A/(1+cot2A)
Take the reciprocal and square roots on both sides, we get
⇒ sec A = ±√ (1+cot2A)/cotA
Now, to express tan function in terms of cot function
tan A = sin A/cos A and cot A = cos A/sin A
Since cot function is the inverse of tan function, it is rewritten as
tan A = 1/cot A

Q2. Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
Cos A function in terms of sec A:
sec A = 1/cos A
⇒ cos A = 1/sec A
sec A function in terms of sec A:
cos2A + sin2A = 1
Rearrange the terms
sin2A = 1 – cos2A
sin2A = 1 – (1/sec2A)
sin2A = (sec2A-1)/sec2A
sin A = ± √(sec2A-1)/sec A
cosec A function in terms of sec A:
sin A = 1/cosec A
⇒cosec A = 1/sin A
cosec A = ± sec A/√(sec2A-1)
Now, tan A function in terms of sec A:
sec2A – tan2A = 1
Rearrange the terms
⇒ tan2A = sec2A + 1
tan A = √(sec2A + 1)
cot A function in terms of sec A:
tan A = 1/cot A
⇒ cot A = 1/tan A
cot A = ±1/√(sec2A + 1)

Q3. Evaluate:
(i) (sin263° + sin227°)/(cos217° + cos273°)
(ii)  sin 25° cos 65° + cos 25° sin 65°
Solution:
(i) (sin263° + sin227°)/(cos217° + cos273°)
To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,
= [sin2(90°-27°) + sin227°] / [cos2(90°-73°) + cos273°)]
= (cos227° + sin227°)/(sin227° + cos273°)
= 1/1 =1                       (since sin2A + cos2A = 1)
Therefore, (sin263° + sin227°)/(cos217° + cos273°) = 1
(ii) sin 25° cos 65° + cos 25° sin 65°
To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,
= sin(90°-25°) cos 65° + cos (90°-65°) sin 65°
= cos 65° cos 65° + sin 65° sin 65°
= cos265° + sin265° = 1 (since sin2A + cos2A = 1)
Therefore, sin 25° cos 65° + cos 25° sin 65° = 1

Q4. 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
Solution:
(i) (B) is correct.
Justification:
Take 9 outside, and it becomes
9 sec2A – 9 tan2A
= 9 (sec2A – tan2A)
= 9×1 = 9             (∵ sec2 A – tan2 A = 1)
Therefore, 9 sec2A – 9 tan2A = 9
(ii) (C) is correct
Justification:
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)
We know that, tan θ = sin θ/cos θ
sec θ = 1/ cos θ
cot θ = cos θ/sin θ
cosec θ = 1/sin θ
Now, substitute the above values in the given problem, we get
= (1 + sin θ/cos θ + 1/ cos θ) (1 + cos θ/sin θ – 1/sin θ)
Simplify the above equation,
= (cos θ +sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ
= (cos θ+sin θ)2-12/(cos θ sin θ)
= (cos2θ + sin2θ + 2cos θ sin θ -1)/(cos θ sin θ)
= (1+ 2cos θ sin θ -1)/(cos θ sin θ) (Since cos2θ + sin2θ = 1)
= (2cos θ sin θ)/(cos θ sin θ) = 2
Therefore, (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =2
(iii) (D) is correct.
Justification:
We know that,
Sec A= 1/cos A
Tan A = sin A / cos A
Now, substitute the above values in the given problem, we get
(secA + tanA) (1 – sinA)
= (1/cos A + sin A/cos A) (1 – sinA)
= (1+sin A/cos A) (1 – sinA)
= (1 – sin2A)/cos A
= cos2A/cos A = cos A
Therefore, (secA + tanA) (1 – sinA) = cos A
(iv) (D) is correct.
Justification:
We know that,
tan2A =1/cot2A
Now, substitute this in the given problem, we get
1+tan2A/1+cot2A
= (1+1/cot2A)/1+cot2A
= (cot2A+1/cot2A)×(1/1+cot2A)
= 1/cot2A = tan2A
So, 1+tan2A/1+cot2A = tan2A


Q5. 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.
NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry (Exercise 8.3)
(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 = tan2
Solution:
(i) (cosec θ – cot θ)= (1-cos θ)/(1+cos θ)
To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)
L.H.S. = (cosec θ – cot θ)2
The above equation is in the form of (a-b)2, and expand it
Since (a-b)2 = a2 + b2 – 2ab
Here a = cosec θ and b = cot θ
= (cosec2θ + cot2θ – 2cosec θ cot θ)
Now, apply the corresponding inverse functions and equivalent ratios to simplify
= (1/sin2θ + cos2θ/sin2θ – 2cos θ/sin2θ)
= (1 + cos2θ – 2cos θ)/(1 – cos2θ)
= (1-cos θ)2/(1 – cosθ)(1+cos θ)
= (1-cos θ)/(1+cos θ) = R.H.S.
Therefore, (cosec θ – cot θ)= (1-cos θ)/(1+cos θ)
Hence proved.
(ii)  (cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A
Now, take the L.H.S of the given equation.
L.H.S. = (cos A/(1+sin A)) + ((1+sin A)/cos A)
= [cos2A + (1+sin A)2]/(1+sin A)cos A
= (cos2A + sin2A + 1 + 2sin A)/(1+sin A) cos A
Since cos2A + sin2A = 1, we can write it as
= (1 + 1 + 2sin A)/(1+sin A) cos A
= (2+ 2sin A)/(1+sin A)cos A
= 2(1+sin A)/(1+sin A)cos A
= 2/cos A = 2 sec A = R.H.S.
L.H.S. = R.H.S.
(cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A
Hence proved.
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)
We know that tan θ =sin θ/cos θ
cot θ = cos θ/sin θ
Now, substitute it in the given equation, to convert it in a simplified form
= [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]
= [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]
= sin2θ/[cos θ(sin θ-cos θ)] + cos2θ/[sin θ(cos θ-sin θ)]
= sin2θ/[cos θ(sin θ-cos θ)] – cos2θ/[sin θ(sin θ-cos θ)]
= 1/(sin θ-cos θ) [(sin2θ/cos θ) – (cos2θ/sin θ)]
= 1/(sin θ-cos θ) × [(sin3θ – cos3θ)/sin θ cos θ]
= [(sin θ-cos θ)(sin2θ+cos2θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]
= (1 + sin θ cos θ)/sin θ cos θ
= 1/sin θ cos θ + 1
= 1 + sec θ cosec θ = R.H.S.
Therefore, L.H.S. = R.H.S.
Hence proved
(iv)  (1 + sec A)/sec A = sin2A/(1-cos A)
First find the simplified form of L.H.S
L.H.S. = (1 + sec A)/sec A
Since secant function is the inverse function of cos function and it is written as
= (1 + 1/cos A)/1/cos A
= (cos A + 1)/cos A/1/cos A
Therefore, (1 + sec A)/sec A = cos A + 1
R.H.S. = sin2A/(1-cos A)
We know that sin2A = (1 – cos2A), we get
= (1 – cos2A)/(1-cos A)
= (1-cos A)(1+cos A)/(1-cos A)
Therefore, sin2A/(1-cos A)= cos A + 1
L.H.S. = R.H.S.
Hence proved
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A, using the identity cosec2A = 1+cot2A.
With the help of identity function, cosec2A = 1+cot2A, let us prove the above equation.
L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)
Divide the numerator and denominator by sin A, we get
= (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A
We know that cos A/sin A = cot A and 1/sin A = cosec A
= (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)
= (cot A – cosec2A + cot2A + cosec A)/(cot A+ 1 – cosec A) (using cosec2A – cot2A = 1
= [(cot A + cosec A) – (cosec2A – cot2A)]/(cot A+ 1 – cosec A)
= [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)
=  (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
=  cot A + cosec A = R.H.S.
Therefore, (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A
Hence Proved
NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry (Exercise 8.3)
First divide the numerator and denominator of L.H.S. by cos A,
NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry (Exercise 8.3)
We know that 1/cos A = sec A and sin A/ cos A = tan A and it becomes,
= √(sec A+ tan A)/(sec A-tan A) 
Now using rationalization, we get
NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry (Exercise 8.3)
= (sec A + tan A)/1
= sec A + tan A = R.H.S
Hence proved
(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
L.H.S. = (sin θ – 2sin3θ)/(2cos3θ – cos θ)
Take sin θ as in numerator and cos θ in denominator as outside, it becomes
= [sin θ(1 – 2sin2θ)]/[cos θ(2cos2θ- 1)]
We know that sin2θ = 1-cos2θ
= sin θ[1 – 2(1-cos2θ)]/[cos θ(2cos2θ -1)]
= [sin θ(2cos2θ -1)]/[cos θ(2cos2θ -1)]
= tan θ = R.H.S.
Hence proved
(viii) (sin A + cosec A)+ (cos A + sec A)2 = 7+tan2A+cot2A
L.H.S. = (sin A + cosec A)+ (cos A + sec A)2
It is of the form (a+b)2, expand it
(a+b)2 =a2 + b2 +2ab
= (sin2A + cosec2A + 2 sin A cosec A) + (cos2A + sec2A + 2 cos A sec A)
= (sin2A + cos2A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan2A + 1 + cot2A
= 1 + 2 + 2 + 2 + tan2A + cot2A
= 7+tan2A+cot2A = R.H.S.
Therefore, (sin A + cosec A)+ (cos A + sec A)2 = 7+tan2A+cot2A
Hence proved.
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A + cotA)
First, find the simplified form of L.H.S
L.H.S. = (cosec A – sin A)(sec A – cos A)
Now, substitute the inverse and equivalent trigonometric ratio forms
= (1/sin A – sin A)(1/cos A – cos A)
= [(1-sin2A)/sin A][(1-cos2A)/cos A]
= (cos2A/sin A)×(sin2A/cos A)
= cos A sin A
Now, simplify the R.H.S
R.H.S. = 1/(tan A+cotA)
= 1/(sin A/cos A +cos A/sin A)
= 1/[(sin2A+cos2A)/sin A cos A]
= cos A sin A
L.H.S. = R.H.S.
(cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
Hence proved
(x)  (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A
L.H.S. = (1+tan2A/1+cot2A)
Since cot function is the inverse of tan function,
= (1+tan2A/1+1/tan2A)
= 1+tan2A/[(1+tan2A)/tan2A]
Now cancel the 1+tan2A terms, we get
= tan2A
(1+tan2A/1+cot2A) = tan2A
Similarly,
(1-tan A/1-cot A)2 = tan2A
Hence proved

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FAQs on NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry (Exercise 8.3)

1. What is trigonometry?
Ans. Trigonometry is a branch of mathematics that deals with the relationships between the angles and sides of triangles. It helps in understanding and calculating various properties of triangles and the measurements of angles.
2. How is trigonometry useful in real life?
Ans. Trigonometry is used in various fields such as engineering, physics, architecture, and navigation. It helps in calculating distances, heights, angles, and positions of objects. For example, it is used in designing buildings, analyzing the motion of objects, and navigating ships and airplanes.
3. What are the basic trigonometric ratios?
Ans. The basic trigonometric ratios are sine, cosine, and tangent. These ratios relate the angles of a right triangle to the lengths of its sides. Sine is the ratio of the length of the side opposite the angle to the hypotenuse, cosine is the ratio of the adjacent side to the hypotenuse, and tangent is the ratio of the opposite side to the adjacent side.
4. How do you find the values of trigonometric ratios?
Ans. The values of trigonometric ratios can be found using a scientific calculator or by referring to trigonometric tables. These tables provide the values of trigonometric ratios for different angles. Alternatively, one can also use the concept of special triangles or the unit circle to find the values of trigonometric ratios for common angles.
5. How can I improve my understanding of trigonometry?
Ans. To improve your understanding of trigonometry, it is important to practice solving various types of trigonometric problems. You can also refer to textbooks, online resources, and video tutorials that explain the concepts in detail. Additionally, seeking help from a teacher or joining a study group can provide additional support and clarification.
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