Class 10  >  Mathematics (Maths) Class 10  >  NCERT Solutions: Introduction to Trigonometry (Exercise 8.4)

NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) Notes | Study Mathematics (Maths) Class 10 - Class 10

Document Description: NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) for Class 10 2022 is part of Mathematics (Maths) Class 10 preparation. The notes and questions for NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) have been prepared according to the Class 10 exam syllabus. Information about NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) covers topics like and NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) Example, for Class 10 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for NCERT Solutions: Introduction to Trigonometry (Exercise 8.4).

Introduction of NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) in English is available as part of our Mathematics (Maths) Class 10 for Class 10 & NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) in Hindi for Mathematics (Maths) Class 10 course. Download more important topics related with notes, lectures and mock test series for Class 10 Exam by signing up for free. Class 10: NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) Notes | Study Mathematics (Maths) Class 10 - Class 10
1 Crore+ students have signed up on EduRev. Have you?

1. 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

2. 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)

3. 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

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
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


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.
NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) Notes | Study Mathematics (Maths) Class 10 - Class 10
(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: Introduction to Trigonometry (Exercise 8.4) Notes | Study Mathematics (Maths) Class 10 - Class 10
First divide the numerator and denominator of L.H.S. by cos A,
NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) Notes | Study Mathematics (Maths) Class 10 - Class 10
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: Introduction to Trigonometry (Exercise 8.4) Notes | Study Mathematics (Maths) Class 10 - Class 10
= (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

The document NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) Notes | Study Mathematics (Maths) Class 10 - Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

Related Searches

video lectures

,

Exam

,

Important questions

,

Free

,

pdf

,

study material

,

practice quizzes

,

Sample Paper

,

Extra Questions

,

MCQs

,

past year papers

,

Previous Year Questions with Solutions

,

mock tests for examination

,

ppt

,

shortcuts and tricks

,

Semester Notes

,

Viva Questions

,

Objective type Questions

,

NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) Notes | Study Mathematics (Maths) Class 10 - Class 10

,

NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) Notes | Study Mathematics (Maths) Class 10 - Class 10

,

NCERT Solutions: Introduction to Trigonometry (Exercise 8.4) Notes | Study Mathematics (Maths) Class 10 - Class 10

,

Summary

;