NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry Exercise 8.4

NCERT Solutions for Class 10 Maths Chapter 8, Introduction to Trigonometry Exercise 8.4, is provided in free PDF format. The Solutions help the students revise all the exercise solutions to score good marks. The NCERT Solutions for Class 10 Maths consists of all solutions with additional information to enhance the learning experience of the students with pictorial illustrations and solutions in a step-by-step procedure.

NCERT solutions are created and analysed as per the guidelines and recently revised syllabus of the CBSE board by the subject matter experts to give correct answers. The solutions are given chapter-wise to access it easily, and students can refer to them anytime for the preparation for the board exams as well as other competitive examinations.

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Access other exercise solutions of Class 10 Maths Chapter 8 – Introduction to Trigonometry

Exercise 8.1 Solutions – 11 Questions (7 short answers, 3 long answers, 1 short answer with reasoning)

Exercise 8.2 Solutions – 4 Questions ( 1 short answer, 2 long answers, 1 MCQ)

Exercise 8.3 Solutions – 7 Questions (5 short answers, 2 long answers)

Access answers of Maths NCERT Class 10 Chapter 8 – Introduction to Trigonometry Exercise 8.4

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 the cot functions, use trigonometric formulas.

We know that,

cosec2A – cot2A = 1

cosec2A = 1 + cot2A

Since the cosec function is the inverse of the 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 the cot function

Now, to express the sec function in terms of the cot function, use the formula

sin2A = 1/ (1+cot2A)

Now, represent the sin function as the cos function

1 – cos2A = 1/ (1+cot2A)

Rearrange the terms,

cos2A = 1 – 1/(1+cot2A)

⇒cos2A = (1-1+cot2A)/(1+cot2A)

Since the sec function is the inverse of the cos function,

⇒ 1/sec2A = cot2A/(1+cot2A)

Take the reciprocal and square roots on both sides, and we get

⇒ sec A = ±√ (1+cot2A)/cotA

Now, to express the tan function in terms of the cot function

tan A = sin A/cos A and cot A = cos A/sin A

Since the cot function is the inverse of the 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 functions into sin functions, and they become,

= [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 functions into sin functions, and they become,

= 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) sec2 A                 (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, and 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, and 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, and 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 L.H.S. and R.H.S. separately]

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

Ncert solutions class 10 chapter 8-10

(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 L.H.S. and R.H.S. separately] (x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A

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 into 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 the secant function is the inverse function of the cos function, 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 the 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, and 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 class 10 chapter 8-11

First, divide the numerator and denominator of L.H.S. by cos A,

Ncert solutions class 10 chapter 8-12

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 rationalisation, we get

Ncert solutions class 10 chapter 8-13

= (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 the 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 the cot function is the inverse of the tan function,

= (1+tan2A/1+1/tan2A)

= 1+tan2A/[(1+tan2A)/tan2A]

Now cancel the 1+tan2A terms, and we get

= tan2A

(1+tan2A/1+cot2A) = tan2A

Similarly,

(1-tan A/1-cot A)2 = tan2A

Hence proved.


Exercise 8.4 of Class 10 Maths recalls the concept of trigonometric identities. In this exercise, a trigonometric identity is defined as an equation which involves the trigonometric ratios of an angle when it is true for all the values of the angle involved in a right-angle triangle. With the help of trigonometric identities, the expression of one trigonometric ratio is written in terms of other trigonometric ratios. By finding one trigonometric identity of the cos and sin functions, the trigonometric identities of other functions are easily found since the functions are interrelated to each other, and the proofs are given elaborately.

In this chapter, trigonometric ratios are expressed in terms of other ratios, proving the trigonometric expressions and justifications with respect to the trigonometric identities given. Also, refer to NCERT solutions for Class 10 maths chapter 8 to solve more problems to score high in the board examinations.

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