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 and helps the students to revise all the exercise solutions completely to score good marks. The NCERT solutions for class 10 maths consists of all solutions with additional information and spice the learning experience of the students with pictorial illustrations and solutions in a step by step procedure.

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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 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) 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, 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 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 LHS and RHS 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 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 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 rationalization, 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 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


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 cos and sin function, the trigonometric identities of other functions are easily found since the functions are inter-related 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 NCERT solutions for class 10 maths chapter 8 to solve more problems to score high in the examinations.

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