# NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

## NCERT Solutions For Class 10 Maths Chapter 8 โ Download Free PDF

**NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry** is helpful for the students as it aids in understanding the concepts as well as in scoring well in CBSE Class 10 board examination. The NCERT Solutions are designed and reviewed by the subject experts that covers all the questions from the textbook. These NCERT Solutions are framed as per the latest CBSE syllabus and guidelines, in accordance with the exam pattern.

The **NCERT Solutions for Class 10 Maths** provides a strong foundation for every concept present across all chapters. Students can clarify their doubts and understand the fundamentals present in this chapter. Also, students can solve the difficult problems in each exercise with the help of these NCERT Solutions for Class 10 Maths Chapter 8.Chapter 8 Introduction to Trigonometry

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Exercise 8.1 Page: 181

**1. In โ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:**

**(i) sin A, cos A**

**(ii) sin C, cos C**

Solution:

In a given triangle ABC, right angled at B = โ B = 90ยฐ

Given: AB = 24 cm and BC = 7 cm

According to the Pythagoras Theorem,

In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

By applying Pythagoras theorem, we get

AC^{2}=AB^{2}+BC^{2}

AC^{2} = (24)^{2}+7^{2}

AC^{2} = (576+49)

AC^{2} = 625cm^{2}

AC = โ625 = 25

Therefore, AC = 25 cm

(i) To find Sin (A), Cos (A)

We know that sine (or) Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side. So it becomes

Sin (A) = Opposite side /Hypotenuse = BC/AC = 7/25

Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes,

Cos (A) = Adjacent side/Hypotenuse = AB/AC = 24/25

(ii) To find Sin (C), Cos (C)

Sin (C) = AB/AC = 24/25

Cos (C) = BC/AC = 7/25

**2. In Fig. 8.13, find tan P โ cot R**

Solution:

In the given triangle PQR, the given triangle is right angled at Q and the given measures are:

PR = 13cm,

PQ = 12cm

Since the given triangle is right angled triangle, to find the side QR, apply the Pythagorean theorem

According to Pythagorean theorem,

In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

PR^{2} = QR^{2} + PQ^{2}

Substitute the values of PR and PQ

13^{2 }= QR^{2}+12^{2}

169 = QR^{2}+144

Therefore, QR^{2 }= 169โ144

QR^{2 }= 25

QR = โ25 = 5

Therefore, the side QR = 5 cm

To find tan P โ cot R:

According to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides, the value of tan (P) becomes

tan (P) **= **Opposite side /Adjacent side = QR/PQ = 5/12

Since cot function is the reciprocal of the tan function, the ratio of cot function becomes,

Cot (R) = Adjacent side/Opposite side = QR/PQ = 5/12

Therefore,

tan (P) โ cot (R) = 5/12 โ 5/12 = 0

Therefore, tan(P) โ cot(R) = 0

**3. If sin A = 3/4, Calculate cos A and tan A.**

Solution:

Let us assume a right angled triangle ABC, right angled at B

Given: Sin A = 3/4

We know that, Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side.

Therefore, Sin A = Opposite side /Hypotenuse= 3/4

Let BC be 3k and AC will be 4k

where k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC^{2}=AB^{2 }+ BC^{2}

Substitute the value of AC and BC

(4k)^{2}=AB^{2} + (3k)^{2}

16k^{2}โ9k^{2 }=AB^{2}

AB^{2}=7k^{2}

Therefore, AB = โ7k

Now, we have to find the value of cos A and tan A

We know that,

Cos (A) = Adjacent side/Hypotenuse

Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get

AB/AC = โ7k/4k = โ7/4

Therefore, cos (A) = โ7/4

tan(A) = Opposite side/Adjacent side

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

BC/AB = 3k/โ7k = 3/โ7

Therefore, tan A = 3/โ7

**4. Given 15 cot A = 8, find sin A and sec A.**

Solution:

Let us assume a right angled triangle ABC, right angled at B

Given: 15 cot A = 8

So, Cot A = 8/15

We know that, cot function is the equal to the ratio of length of the adjacent side to the opposite side.

Therefore, cot A = Adjacent side/Opposite side = AB/BC = 8/15

Let AB be 8k and BC will be 15k

Where, k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC^{2}=AB^{2 }+ BC^{2}

Substitute the value of AB and BC

AC^{2}= (8k)^{2} + (15k)^{2}

AC^{2}= 64k^{2} + 225k^{2}

AC^{2}= 289k^{2}

Therefore, AC = 17k

Now, we have to find the value of sin A and sec A

We know that,

Sin (A) = Opposite side /Hypotenuse

Substitute the value of BC and AC and cancel the constant k in both numerator and denominator, we get

Sin A = BC/AC = 15k/17k = 15/17

Therefore, sin A = 15/17

Since secant or sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side.

Sec (A) = Hypotenuse/Adjacent side

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

AC/AB = 17k/8k = 17/8

Therefore sec (A) = 17/8

**5. Given sec ฮธ = 13/12 Calculate all other trigonometric ratios**

Solution:

We know that sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side

Let us assume a right angled triangle ABC, right angled at B

sec ฮธ =13/12 = Hypotenuse/Adjacent side = AC/AB

Let AC be 13k and AB will be 12k

Where, k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC^{2}=AB^{2 }+ BC^{2}

Substitute the value of AB and AC

(13k)^{2}= (12k)^{2} + BC^{2}

169k^{2}= 144k^{2} + BC^{2}

169k^{2}= 144k^{2} + BC^{2}

BC^{2 = }169k^{2} โ 144k^{2}

BC^{2}= 25k^{2}

Therefore, BC = 5k

Now, substitute the corresponding values in all other trigonometric ratios

So,

Sin ฮธ = Opposite Side/Hypotenuse = BC/AC = 5/13

Cos ฮธ = Adjacent Side/Hypotenuse = AB/AC = 12/13

tan ฮธ = Opposite Side/Adjacent Side = BC/AB = 5/12

Cosec ฮธ = Hypotenuse/Opposite Side = AC/BC = 13/5

cot ฮธ = Adjacent Side/Opposite Side = AB/BC = 12/5

**6. If โ A and โ B are acute angles such that cos A = cos B, then show that โ A = โ B.**

Solution:

Let us assume the triangle ABC in which CDโฅAB

Give that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cos ratio is written as

AD/AC = BD/BC

Now, interchange the terms, we get

AD/BD = AC/BC

Let take a constant value

AD/BD = AC/BC = k

Now consider the equation as

AD = k BD โฆ(1)

AC = k BC โฆ(2)

By applying Pythagoras theorem in โณCAD and โณCBD we get,

CD^{2} = BC^{2} โ BD^{2 }โฆ (3)

CD^{2 }=AC^{2 }โAD^{2} โฆ.(4)

From the equations (3) and (4) we get,

AC^{2}โAD^{2 }= BC^{2}โBD^{2}

Now substitute the equations (1) and (2) in (3) and (4)

K^{2}(BC^{2}โBD^{2})=(BC^{2}โBD^{2}) k^{2}=1

Putting this value in equation, we obtain

AC = BC

โ A=โ B (Angles opposite to equal side are equal-isosceles triangle)

**7. If cot ฮธ = 7/8, evaluate :**

**(i) (1 + sin ฮธ)(1 โ sin ฮธ)/(1+cos ฮธ)(1-cos ฮธ)**

**(ii) cot ^{2} ฮธ**

Solution:

Let us assume a โณABC in which โ B = 90ยฐ and โ C = ฮธ

Given:

cot ฮธ = BC/AB = 7/8

Let BC = 7k and AB = 8k, where k is a positive real number

According to Pythagoras theorem in โณABC we get.

AC^{2 }= AB^{2}+BC^{2}

AC^{2 }= (8k)^{2}+(7k)^{2}

AC^{2 }= 64k^{2}+49k^{2}

AC^{2 }= 113k^{2}

AC = โ113 k

According to the sine and cos function ratios, it is written as

sin ฮธ = AB/AC = Opposite Side/Hypotenuse = 8k/โ113 k = 8/โ113 and

cos ฮธ = Adjacent Side/Hypotenuse = BC/AC = 7k/โ113 k = 7/โ113

Now apply the values of sin function and cos function:

**8. If 3 cot A = 4, check whether (1-tan ^{2 }A)/(1+tan^{2} A) = cos^{2} A โ sin ^{2 }A or not.**

Solution:

Let โณABC in which โ B=90ยฐ

We know that, cot function is the reciprocal of tan function and it is written as

cot(A) = AB/BC = 4/3

Let AB = 4k an BC =3k, where k is a positive real number.

According to the Pythagorean theorem,

AC^{2}=AB^{2}+BC^{2}

AC^{2}=(4k)^{2}+(3k)^{2}

AC^{2}=16k^{2}+9k^{2}

AC^{2}=25k^{2}

AC=5k

Now, apply the values corresponding to the ratios

tan(A) = BC/AB = 3/4

sin (A) = BC/AC = 3/5

cos (A) = AB/AC = 4/5

Now compare the left hand side(LHS) with right hand side(RHS)

Since, both the LHS and RHS = 7/25

R.H.S. =L.H.S.

Hence, **(1-tan ^{2 }A)/(1+tan^{2} A) = cos^{2} A โ sin ^{2 }A** is proved

**9. In triangle ABC, right-angled at B, if tan A = 1/โ3 find the value of:**

**(i) sin A cos C + cos A sin C**

**(ii) cos A cos C โ sin A sin C**

Solution:

Let ฮABC in which โ B=90ยฐ

tan A = BC/AB = 1/โ3

Let BC = 1k and AB = โ3 k,

Where k is the positive real number of the problem

By Pythagoras theorem in ฮABC we get:

AC^{2}=AB^{2}+BC^{2}

AC^{2}=(โ3 k)^{2}+(k)^{2}

AC^{2}=3k^{2}+k^{2}

AC^{2}=4k^{2}

AC = 2k

Now find the values of cos A, Sin A

Sin A = BC/AC = 1/2

Cos A = AB/AC = โ3/2

Then find the values of cos C and sin C

Sin C = AB/AC = **โ**3/2

Cos C = BC/AC = 1/2

Now, substitute the values in the given problem

(i) sin A cos C + cos A sin C = (1/2) ร(1/2 )+ โ3/2 รโ3/2 = 1/4 + 3/4 = 1

(ii) cos A cos C โ sin A sin C = (**โ**3/2 )(1/2) โ (1/2) (**โ**3/2 ) = 0

**10. In โ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P**

Solution:

In a given triangle PQR, right angled at Q, the following measures are

PQ = 5 cm

PR + QR = 25 cm

Now let us assume, QR = x

PR = 25-QR

PR = 25- x

According to the Pythagorean Theorem,

PR^{2} = PQ^{2} + QR^{2}

Substitute the value of PR as x

(25- x)^{ 2 }= 5^{2 }+ x^{2}

25^{2} + x^{2} โ 50x = 25 + x^{2}

625 + x^{2}-50x -25 โ x^{2 }= 0

-50x = -600

x= -600/-50

x = 12 = QR

Now, find the value of PR

PR = 25- QR

Substitute the value of QR

PR = 25-12

PR = 13

Now, substitute the value to the given problem

(1) sin p = Opposite Side/Hypotenuse = QR/PR = 12/13

(2) Cos p = Adjacent Side/Hypotenuse = PQ/PR = 5/13

(3) tan p =Opposite Side/Adjacent side = QR/PQ = 12/5

**11. State whether the following are true or false. Justify your answer.**

**(i) The value of tan A is always less than 1.**

**(ii) sec A = 12/5 for some value of angle A.**

**(iii)cos A is the abbreviation used for the cosecant of angle A.**

**(iv) cot A is the product of cot and A.**

**(v) sin ฮธ = 4/3 for some angle ฮธ.**

Solution:

**(i) **The value of tan A is always less than 1.

Answer: **False**

Proof: In ฮMNC in which โ N = 90โ,

MN = 3, NC = 4 and MC = 5

Value of tan M = 4/3 which is greater than.

The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem.

MC^{2}=MN^{2}+NC^{2}

5^{2}=3^{2}+4^{2}

25=9+16

25^{ }=^{ }25

**(ii)** sec A = 12/5 for some value of angle A

Answer: **True**

Justification: Let a ฮMNC in which โ N = 90ยบ,

MC=12k and MB=5k, where k is a positive real number.

By Pythagoras theorem we get,

MC^{2}=MN^{2}+NC^{2}

(12k)^{2}=(5k)^{2}+NC^{2}

NC^{2}+25k^{2}=144k^{2}

NC^{2}=119k^{2}

Such a triangle is possible as it will follow the Pythagoras theorem.

**(iii)** cos A is the abbreviation used for the cosecant of angle A.

Answer: **False**

Justification: Abbreviation used for cosecant of angle M is cosec M. cos M is the abbreviation used for cosine of angle M.

**(iv)** cot A is the product of cot and A.

Answer:** False**

Justification: cot M is not the product of cot and M. It is the cotangent of โ M.

**(v)** sin ฮธ = 4/3 for some angle ฮธ.

Answer**: False**

Justification: sin ฮธ = Height/Hypotenuse

We know that in a right angled triangle, Hypotenuse is the longest side.

โด sin ฮธ will always less than 1 and it can never be 4/3 for any value of ฮธ.

Exercise 8.2 Page: 187

**1. Evaluate the following:**

**(i) sin 60ยฐ cos 30ยฐ + sin 30ยฐ cos 60ยฐ**

**(ii) 2 tan ^{2} 45ยฐ + cos^{2} 30ยฐ โ sin^{2} 60**

Solution:

(i) sin 60ยฐ cos 30ยฐ + sin 30ยฐ cos 60ยฐ

First, find the values of the given trigonometric ratios

sin 30ยฐ = 1/2

cos 30ยฐ = โ3/2

sin 60ยฐ = 3/2

cos 60ยฐ= 1/2

Now, substitute the values in the given problem

sin 60ยฐ cos 30ยฐ + sin 30ยฐ cos 60ยฐ = โ3/2 รโ3/2 + (1/2) ร(1/2 ) = 3/4+1/4 = 4/4 =

(ii) 2 tan^{2} 45ยฐ + cos^{2} 30ยฐ โ sin^{2} 60

We know that, the values of the trigonometric ratios are:

sin 60ยฐ = โ3/2

cos 30ยฐ = โ3/2

tan 45ยฐ = 1

Substitute the values in the given problem

2 tan^{2} 45ยฐ + cos^{2} 30ยฐ โ sin^{2} 60 = 2(1)^{2 }+ (โ3/2)^{2}-(โ3/2)^{2}

2 tan^{2} 45ยฐ + cos^{2} 30ยฐ โ sin^{2} 60 = 2 + 0

2 tan^{2} 45ยฐ + cos^{2} 30ยฐ โ sin^{2} 60 = 2

(iii) cos 45ยฐ/(sec 30ยฐ+cosec 30ยฐ)

We know that,

cos 45ยฐ = 1/โ2

sec 30ยฐ = 2/โ3

cosec 30ยฐ = 2

Substitute the values, we get

Now, multiply both the numerator and denominator by โ2 , we get

Therefore, cos 45ยฐ/(sec 30ยฐ+cosec 30ยฐ) = (3โ2 โ โ6)/8

We know that,

sin 30ยฐ = 1/2

tan 45ยฐ = 1

cosec 60ยฐ = 2/โ3

sec 30ยฐ = 2/โ3

cos 60ยฐ = 1/2

cot 45ยฐ = 1

Substitute the values in the given problem, we get

We know that,

cos 60ยฐ = 1/2

sec 30ยฐ = 2/โ3

tan 45ยฐ = 1

sin 30ยฐ = 1/2

cos 30ยฐ = โ3/2

Now, substitute the values in the given problem, we get

(5cos^{2}60ยฐ + 4sec^{2}30ยฐ โ tan^{2}45ยฐ)/(sin^{2 }30ยฐ + cos^{2 }30ยฐ)

= 5(1/2)^{2}+4(2/โ3)^{2}-1^{2}/(1/2)^{2}+(โ3/2)^{2}

^{ }= (5/4+16/3-1)/(1/4+3/4)

= (15+64-12)/12/(4/4)

= 67/12

**2. Choose the correct option and justify your choice :(i) 2tan 30ยฐ/1+tan ^{2}30ยฐ =(A) sin 60ยฐ (B) cos 60ยฐ (C) tan 60ยฐ (D) sin 30ยฐ(ii) 1-tan^{2}45ยฐ/1+tan^{2}45ยฐ =(A) tan 90ยฐ (B) 1 (C) sin 45ยฐ (D) 0(iii) sin 2A = 2 sin A is true when A =(A) 0ยฐ (B) 30ยฐ (C) 45ยฐ (D) 60ยฐ**

**(iv) 2tan30ยฐ/1-tan ^{2}30ยฐ =(A) cos 60ยฐ (B) sin 60ยฐ (C) tan 60ยฐ (D) sin 30ยฐ**

Solution:

(i) (A) is correct.

Substitute the of tan 30ยฐ in the given equation

tan 30ยฐ = 1/โ3

2tan 30ยฐ/1+tan^{2}30ยฐ = 2(1/โ3)/1+(1/โ3)^{2}

= (2/โ3)/(1+1/3) = (2/โ3)/(4/3)

= 6/4โ3 = โ3/2 = sin 60ยฐ

The obtained solution is equivalent to the trigonometric ratio sin 60ยฐ

(ii) (D) is correct.

Substitute the of tan 45ยฐ in the given equation

tan 45ยฐ = 1

1-tan^{2}45ยฐ/1+tan^{2}45ยฐ = (1-1^{2})/(1+1^{2})

= 0/2 = 0

The solution of the above equation is 0.

(iii) (A) is correct.

To find the value of A, substitute the degree given in the options one by one

sin 2A = 2 sin A is true when A = 0ยฐ

As sin 2A = sin 0ยฐ = 0

2 sin A = 2 sin 0ยฐ = 2 ร 0 = 0

or,

Apply the sin 2A formula, to find the degree value

sin 2A = 2sin A cos A

โ2sin A cos A = 2 sin A

โ 2cos A = 2 โ cos A = 1

Now, we have to check, to get the solution as 1, which degree value has to be applied.

When 0 degree is applied to cos value, i.e., cos 0 =1

Therefore, โ A = 0ยฐ

(iv) (C) is correct.

Substitute the of tan 30ยฐ in the given equation

tan 30ยฐ = 1/โ3

2tan30ยฐ/1-tan^{2}30ยฐ = 2(1/โ3)/1-(1/โ3)^{2}

= (2/โ3)/(1-1/3) = (2/โ3)/(2/3) = โ3 = tan 60ยฐ

The value of the given equation is equivalent to tan 60ยฐ.

**3. If tan (A + B) = โ3 and tan (A โ B) = 1/โ3 ,0ยฐ < A + B โค 90ยฐ; A > B, find A and B.**

Solution:

tan (A + B) = โ3

Since โ3 = tan 60ยฐ

Now substitute the degree value

โ tan (A + B) = tan 60ยฐ

(A + B) = 60ยฐ โฆ (i)

The above equation is assumed as equation (i)

tan (A โ B) = 1/โ3

Since 1/โ3 = tan 30ยฐ

Now substitute the degree value

โ tan (A โ B) = tan 30ยฐ

(A โ B) = 30ยฐ โฆ equation (ii)

Now add the equation (i) and (ii), we get

A + B + A โ B = 60ยฐ + 30ยฐ

Cancel the terms B

2A = 90ยฐ

A= 45ยฐ

Now, substitute the value of A in equation (i) to find the value of B

45ยฐ + B = 60ยฐ

B = 60ยฐ โ 45ยฐ

B = 15ยฐ

Therefore A = 45ยฐ and B = 15ยฐ

**4. State whether the following are true or false. Justify your answer.**

**(i) sin (A + B) = sin A + sin B.**

**(ii) The value of sin ฮธ increases as ฮธ increases.**

**(iii) The value of cos ฮธ increases as ฮธ increases.**

**(iv) sin ฮธ = cos ฮธ for all values of ฮธ.**

**(v) cot A is not defined for A = 0ยฐ.**

Solution:

(i) False.

Justification:

Let us take A = 30ยฐ and B = 60ยฐ, then

Substitute the values in the sin (A + B) formula, we get

sin (A + B) = sin (30ยฐ + 60ยฐ) = sin 90ยฐ = 1 and,

sin A + sin B = sin 30ยฐ + sin 60ยฐ

= 1/2 + โ3/2 = 1+โ3/2

Since the values obtained are not equal, the solution is false.

(ii) True.

Justification:

According to the values obtained as per the unit circle, the values of sin are:

sin 0ยฐ = 0

sin 30ยฐ = 1/2

sin 45ยฐ = 1/โ2

sin 60ยฐ = โ3/2

sin 90ยฐ = 1

Thus the value of sin ฮธ increases as ฮธ increases. Hence, the statement is true

(iii) False.

According to the values obtained as per the unit circle, the values of cos are:

cos 0ยฐ = 1

cos 30ยฐ = โ3/2

cos 45ยฐ = 1/โ2

cos 60ยฐ = 1/2

cos 90ยฐ = 0

Thus, the value of cos ฮธ decreases as ฮธ increases. So, the statement given above is false.

(iv) False

sin ฮธ = cos ฮธ, when a right triangle has 2 angles of (ฯ/4). Therefore, the above statement is false.

(v) True.

Since cot function is the reciprocal of the tan function, it is also written as:

cot A = cos A/sin A

Now substitute A = 0ยฐ

cot 0ยฐ = cos 0ยฐ/sin 0ยฐ = 1/0 = undefined.

Hence, it is true

Exercise 8.3 Page: 189

**1. Evaluate :**

**(i) sin 18ยฐ/cos 72ยฐ **

**(ii) tan 26ยฐ/cot 64ยฐ **

**(iii) cos 48ยฐ โ sin 42ยฐ **

**(iv) cosec 31ยฐ โ sec 59ยฐ**

Solution:

(i) sin 18ยฐ/cos 72ยฐ

To simplify this, convert the sin function into cos function

We know that, 18ยฐ is written as 90ยฐ โ 18ยฐ, which is equal to the cos 72ยฐ.

= sin (90ยฐ โ 18ยฐ) /cos 72ยฐ

Substitute the value, to simplify this equation

= cos 72ยฐ /cos 72ยฐ = 1

(ii) tan 26ยฐ/cot 64ยฐ

To simplify this, convert the tan function into cot function

We know that, 26ยฐ is written as 90ยฐ โ 36ยฐ, which is equal to the cot 64ยฐ.

= tan (90ยฐ โ 36ยฐ)/cot 64ยฐ

Substitute the value, to simplify this equation

= cot 64ยฐ/cot 64ยฐ = 1

(iii) cos 48ยฐ โ sin 42ยฐ

To simplify this, convert the cos function into sin function

We know that, 48ยฐ is written as 90ยฐ โ 42ยฐ, which is equal to the sin 42ยฐ.

= cos (90ยฐ โ 42ยฐ) โ sin 42ยฐ

Substitute the value, to simplify this equation

= sin 42ยฐ โ sin 42ยฐ = 0

(iv) cosec 31ยฐ โ sec 59ยฐ

To simplify this, convert the cosec function into sec function

We know that, 31ยฐ is written as 90ยฐ โ 59ยฐ, which is equal to the sec 59ยฐ

= cosec (90ยฐ โ 59ยฐ) โ sec 59ยฐ

Substitute the value, to simplify this equation

= sec 59ยฐ โ sec 59ยฐ = 0

**2. Show that:**

**(i) tan 48ยฐ tan 23ยฐ tan 42ยฐ tan 67ยฐ = 1**

**(ii) cos 38ยฐ cos 52ยฐ โ sin 38ยฐ sin 52ยฐ = 0**

Solution:

(i) tan 48ยฐ tan 23ยฐ tan 42ยฐ tan 67ยฐ

Simplify the given problem by converting some of the tan functions to the cot functions

We know that, tan 48ยฐ = tan (90ยฐ โ 42ยฐ) = cot 42ยฐ

tan 23ยฐ = tan (90ยฐ โ 67ยฐ) = cot 67ยฐ

= tan (90ยฐ โ 42ยฐ) tan (90ยฐ โ 67ยฐ) tan 42ยฐ tan 67ยฐ

Substitute the values

= cot 42ยฐ cot 67ยฐ tan 42ยฐ tan 67ยฐ

= (cot 42ยฐ tan 42ยฐ) (cot 67ยฐ tan 67ยฐ) = 1ร1 = 1

(ii) cos 38ยฐ cos 52ยฐ โ sin 38ยฐ sin 52ยฐ

Simplify the given problem by converting some of the cos functions to the sin functions

We know that,

cos 38ยฐ = cos (90ยฐ โ 52ยฐ) = sin 52ยฐ

cos 52ยฐ= cos (90ยฐ-38ยฐ) = sin 38ยฐ

= cos (90ยฐ โ 52ยฐ) cos (90ยฐ-38ยฐ) โ sin 38ยฐ sin 52ยฐ

Substitute the values

= sin 52ยฐ sin 38ยฐ โ sin 38ยฐ sin 52ยฐ = 0

**3. If tan 2A = cot (A โ 18ยฐ), where 2A is an acute angle, find the value of A**.

Solution:

tan 2A = cot (A- 18ยฐ)

We know that tan 2A = cot (90ยฐ โ 2A)

Substitute the above equation in the given problem

โ cot (90ยฐ โ 2A) = cot (A -18ยฐ)

Now, equate the angles,

โ 90ยฐ โ 2A = A- 18ยฐ โ 108ยฐ = 3A

A = 108ยฐ / 3

Therefore, the value of A = 36ยฐ

**4. If tan A = cot B, prove that A + B = 90ยฐ.**

Solution:

tan A = cot B

We know that cot B = tan (90ยฐ โ B)

To prove A + B = 90ยฐ, substitute the above equation in the given problem

tan A = tan (90ยฐ โ B)

A = 90ยฐ โ B

A + B = 90ยฐ

Hence Proved.

**5. If sec 4A = cosec (A โ 20ยฐ), where 4A is an acute angle, find the value of A.**

Solution:

sec 4A = cosec (A โ 20ยฐ)

We know that sec 4A = cosec (90ยฐ โ 4A)

To find the value of A, substitute the above equation in the given problem

cosec (90ยฐ โ 4A) = cosec (A โ 20ยฐ)

Now, equate the angles

90ยฐ โ 4A= A- 20ยฐ

110ยฐ = 5A

A = 110ยฐ/ 5 = 22ยฐ

Therefore, the value of A = 22ยฐ

**6. If A, B and C are interior angles of a triangle ABC, then show that**

** sin (B+C/2) = cos A/2**

Solution:

We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180ยฐ

A + B + C = 180ยฐ โฆ.(1)

To find the value of (B+ C)/2, simplify the equation (1)

โ B + C = 180ยฐ โ A

โ (B+C)/2 = (180ยฐ-A)/2

โ (B+C)/2 = (90ยฐ-A/2)

Now, multiply both sides by sin functions, we get

โ sin (B+C)/2 = sin (90ยฐ-A/2)

Since sin (90ยฐ-A/2) = cos A/2, the above equation is equal to

sin (B+C)/2 = cos A/2

Hence proved.

**7. Express sin 67ยฐ + cos 75ยฐ in terms of trigonometric ratios of angles between 0ยฐ and 45ยฐ.**

Solution:

Given:

sin 67ยฐ + cos 75ยฐ

In term of sin as cos function and cos as sin function, it can be written as follows

sin 67ยฐ = sin (90ยฐ โ 23ยฐ)

cos 75ยฐ = cos (90ยฐ โ 15ยฐ)

So, sin 67ยฐ + cos 75ยฐ = sin (90ยฐ โ 23ยฐ) + cos (90ยฐ โ 15ยฐ)

Now, simplify the above equation

= cos 23ยฐ + sin 15ยฐ

Therefore, sin 67ยฐ + cos 75ยฐ is also expressed as cos 23ยฐ + sin 15ยฐ

Exercise 8.4 Page: 193

**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,

cosec^{2}A^{ }โ cot^{2}A = 1

cosec^{2}A = 1 + cot^{2}A

Since cosec function is the inverse of sin function, it is written as

1/sin^{2}A = 1 + cot^{2}A

Now, rearrange the terms, it becomes

sin^{2}A = 1/(1+cot^{2}A)

Now, take square roots on both sides, we get

sin A = ยฑ1/(โ(1+cot^{2}A)

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

sin^{2}A = 1/ (1+cot^{2}A)

Now, represent the sin function as cos function

1 โ cos^{2}A = 1/ (1+cot^{2}A)

Rearrange the terms,

cos^{2}A = 1 โ 1/(1+cot^{2}A)

โcos^{2}A = (1-1+cot^{2}A)/(1+cot^{2}A)

Since sec function is the inverse of cos function,

โ 1/sec^{2}A = cot^{2}A/(1+cot^{2}A)

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

โ sec A = ยฑโ (1+cot^{2}A)/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 โ A in terms of sec A.**

Solution:

Cos A function in terms of sec A:

sec A = 1/cos A

โ cos A = 1/sec A

sec A function in terms of sec A:

cos^{2}A + sin^{2}A = 1

Rearrange the terms

sin^{2}A = 1 โ cos^{2}A

sin^{2}A = 1 โ (1/sec^{2}A)

sin^{2}A = (sec^{2}A-1)/sec^{2}A

sin A = ยฑ โ(sec^{2}A-1)/sec A

cosec A function in terms of sec A:

sin A = 1/cosec A

โcosec A = 1/sin A

cosec A = ยฑ sec A/โ(sec^{2}A-1)

Now, tan A function in terms of sec A:

sec^{2}A โ tan^{2}A = 1

Rearrange the terms

โ tan^{2}A = sec^{2}A โ 1

tan A = โ(sec^{2}A โ 1)

cot A function in terms of sec A:

tan A = 1/cot A

โ cot A = 1/tan A

cot A = ยฑ1/โ(sec^{2}A โ 1)

**3. Evaluate:**

**(i) (sin ^{2}63ยฐ + sin^{2}27ยฐ)/(cos^{2}17ยฐ + cos^{2}73ยฐ)**

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

Solution:

(i) (sin^{2}63ยฐ + sin^{2}27ยฐ)/(cos^{2}17ยฐ + cos^{2}73ยฐ)

To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,

= [sin^{2}(90ยฐ-27ยฐ) + sin^{2}27ยฐ] / [cos^{2}(90ยฐ-73ยฐ) + cos^{2}73ยฐ)]

= (cos^{2}27ยฐ^{ }+ sin^{2}27ยฐ)/(sin^{2}27ยฐ + cos^{2}73ยฐ)

= 1/1 =1 (since sin^{2}A + cos^{2}A = 1)

Therefore, (sin^{2}63ยฐ + sin^{2}27ยฐ)/(cos^{2}17ยฐ + cos^{2}73ยฐ) = 1

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

To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,

= sin(90ยฐ-25ยฐ) cos 65ยฐ + cos (90ยฐ-65ยฐ) sin 65ยฐ

= cos 65ยฐ cos 65ยฐ + sin 65ยฐ sin 65ยฐ

= cos^{2}65ยฐ^{ }+ sin^{2}65ยฐ = 1 (since sin^{2}A + cos^{2}A = 1)

Therefore, sin 25ยฐ cos 65ยฐ + cos 25ยฐ sin 65ยฐ = 1

**4. Choose the correct option. Justify your choice.(i) 9 sec ^{2}A โ 9 tan^{2}A =(A) 1 (B) 9 (C) 8 (D) 0(ii) (1 + tan ฮธ + sec ฮธ) (1 + cot ฮธ โ cosec ฮธ)(A) 0 (B) 1 (C) 2 (D) โ 1(iii) (sec A + tan A) (1 โ sin A) =(A) sec A (B) sin A (C) cosec A (D) cos A**

**(iv) 1+tan ^{2}A/1+cot^{2}A = **

** (A) sec ^{2 }A (B) -1 (C) cot^{2}A (D) tan^{2}A**

Solution:

(i) (B) is correct.

Justification:

Take 9 outside, and it becomes

9 sec^{2}A โ 9 tan^{2}A

= 9 (sec^{2}A โ tan^{2}A)

= 9ร1 = 9 (โต sec2 A โ tan2 A = 1)

Therefore, 9 sec^{2}A โ 9 tan^{2}A = 9

(ii) (C) is correct

Justification:

(1 + tan ฮธ + sec ฮธ) (1 + cot ฮธ โ cosec ฮธ)

We know that, tan ฮธ = sin ฮธ/cos ฮธ

sec ฮธ = 1/ cos ฮธ

cot ฮธ = cos ฮธ/sin ฮธ

cosec ฮธ = 1/sin ฮธ

Now, substitute the above values in the given problem, we get

= (1 + sin ฮธ/cos ฮธ + 1/ cos ฮธ) (1 + cos ฮธ/sin ฮธ โ 1/sin ฮธ)

Simplify the above equation,

= (cos ฮธ +sin ฮธ+1)/cos ฮธ ร (sin ฮธ+cos ฮธ-1)/sin ฮธ

= (cos ฮธ+sin ฮธ)^{2}-1^{2}/(cos ฮธ sin ฮธ)

= (cos^{2}ฮธ + sin^{2}ฮธ + 2cos ฮธ sin ฮธ -1)/(cos ฮธ sin ฮธ)

= (1+ 2cos ฮธ sin ฮธ -1)/(cos ฮธ sin ฮธ) (Since cos^{2}ฮธ + sin^{2}ฮธ = 1)

= (2cos ฮธ sin ฮธ)/(cos ฮธ sin ฮธ) = 2

Therefore, (1 + tan ฮธ + sec ฮธ) (1 + cot ฮธ โ cosec ฮธ) =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 โ sinA)

= (1/cos A + sin A/cos A) (1 โ sinA)

= (1+sin A/cos A) (1 โ sinA)

= (1 โ sin^{2}A)/cos A

= cos^{2}A/cos A = cos A

Therefore, (secA + tanA) (1 โ sinA) = cos A

(iv) (D) is correct.

Justification:

We know that,

tan^{2}A =1/cot^{2}A

Now, substitute this in the given problem, we get

1+tan^{2}A/1+cot^{2}A

= (1+1/cot^{2}A)/1+cot^{2}A

= (cot^{2}A+1/cot^{2}A)ร(1/1+cot^{2}A)

= 1/cot^{2}A = tan^{2}A

So, 1+tan^{2}A/1+cot^{2}A = tan^{2}A

**5. Prove the following identities, where the angles involved are acute angles for which theexpressions are defined.**

**(i) (cosec ฮธ โ cot ฮธ) ^{2 }= (1-cos ฮธ)/(1+cos ฮธ)**

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

**(iii) tan ฮธ/(1-cot ฮธ) + cot ฮธ/(1-tan ฮธ) = 1 + sec ฮธ cosec ฮธ**

** [Hint : Write the expression in terms of sin ฮธ and cos ฮธ]**

**(iv) (1 + sec A)/sec A = sin ^{2}A/(1-cos A) **

** [Hint : Simplify LHS and RHS separately]**

**(v) ( cos Aโsin A+1)/( cos A +sin Aโ1) = cosec A + cot A, using the identity cosec ^{2}A = 1+cot^{2}A.**

**(vii) (sin ฮธ โ 2sin ^{3}ฮธ)/(2cos^{3}ฮธ-cos ฮธ) = tan ฮธ**

(viii) (sin A + cosec A)^{2 }+ (cos A + sec A)^{2} = 7+tan^{2}A+cot^{2}A

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

[Hint : Simplify LHS and RHS separately]

(x) (1+tan^{2}A/1+cot^{2}A) = (1-tan A/1-cot A)^{2} =^{ }tan^{2}A

Solution:

(i) (cosec ฮธ โ cot ฮธ)^{2 }= (1-cos ฮธ)/(1+cos ฮธ)

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 ฮธ โ cot ฮธ)^{2}

The above equation is in the form of (a-b)^{2}, and expand it

Since (a-b)^{2} = a^{2} + b^{2} โ 2ab

Here a = cosec ฮธ and b = cot ฮธ

= (cosec^{2}ฮธ + cot^{2}ฮธ โ 2cosec ฮธ cot ฮธ)

Now, apply the corresponding inverse functions and equivalent ratios to simplify

= (1/sin^{2}ฮธ + cos^{2}ฮธ/sin^{2}ฮธ โ 2cos ฮธ/sin^{2}ฮธ)

= (1 + cos^{2}ฮธ โ 2cos ฮธ)/(1 โ cos^{2}ฮธ)

= (1-cos ฮธ)^{2}/(1 โ cosฮธ)(1+cos ฮธ)

= (1-cos ฮธ)/(1+cos ฮธ) = R.H.S.

Therefore, (cosec ฮธ โ cot ฮธ)^{2 }= (1-cos ฮธ)/(1+cos ฮธ)

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)

= [cos^{2}A + (1+sin A)^{2}]/(1+sin A)cos A

= (cos^{2}A + sin^{2}A + 1 + 2sin A)/(1+sin A) cos A

Since cos^{2}A + sin^{2}A = 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 ฮธ/(1-cot ฮธ) + cot ฮธ/(1-tan ฮธ) = 1 + sec ฮธ cosec ฮธ

L.H.S. = tan ฮธ/(1-cot ฮธ) + cot ฮธ/(1-tan ฮธ)

We know that tan ฮธ =sin ฮธ/cos ฮธ

cot ฮธ = cos ฮธ/sin ฮธ

Now, substitute it in the given equation, to convert it in a simplified form

= [(sin ฮธ/cos ฮธ)/1-(cos ฮธ/sin ฮธ)] + [(cos ฮธ/sin ฮธ)/1-(sin ฮธ/cos ฮธ)]

= [(sin ฮธ/cos ฮธ)/(sin ฮธ-cos ฮธ)/sin ฮธ] + [(cos ฮธ/sin ฮธ)/(cos ฮธ-sin ฮธ)/cos ฮธ]

= sin^{2}ฮธ/[cos ฮธ(sin ฮธ-cos ฮธ)] + cos^{2}ฮธ/[sin ฮธ(cos ฮธ-sin ฮธ)]

= sin^{2}ฮธ/[cos ฮธ(sin ฮธ-cos ฮธ)] โ cos^{2}ฮธ/[sin ฮธ(sin ฮธ-cos ฮธ)]

= 1/(sin ฮธ-cos ฮธ) [(sin^{2}ฮธ/cos ฮธ) โ (cos^{2}ฮธ/sin ฮธ)]

= 1/(sin ฮธ-cos ฮธ) ร [(sin^{3}ฮธ โ cos^{3}ฮธ)/sin ฮธ cos ฮธ]

= [(sin ฮธ-cos ฮธ)(sin^{2}ฮธ+cos^{2}ฮธ+sin ฮธ cos ฮธ)]/[(sin ฮธ-cos ฮธ)sin ฮธ cos ฮธ]

= (1 + sin ฮธ cos ฮธ)/sin ฮธ cos ฮธ

= 1/sin ฮธ cos ฮธ + 1

= 1 + sec ฮธ cosec ฮธ = R.H.S.

Therefore, L.H.S. = R.H.S.

Hence proved

(iv) (1 + sec A)/sec A = sin^{2}A/(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. = sin^{2}A/(1-cos A)

We know that sin^{2}A = (1 โ cos^{2}A), we get

= (1 โ cos^{2}A)/(1-cos A)

= (1-cos A)(1+cos A)/(1-cos A)

Therefore, sin^{2}A/(1-cos A)= cos A + 1

L.H.S. = R.H.S.

Hence proved

(v) (cos Aโsin A+1)/(cos A+sin Aโ1) = cosec A + cot A, using the identity cosec^{2}A = 1+cot^{2}A.

With the help of identity function, cosec^{2}A = 1+cot^{2}A, let us prove the above equation.

L.H.S. = (cos Aโsin A+1)/(cos A+sin Aโ1)

Divide the numerator and denominator by sin A, we get

= (cos Aโsin A+1)/sin A/(cos A+sin Aโ1)/sin A

We know that cos A/sin A = cot A and 1/sin A = cosec A

= (cot A โ 1 + cosec A)/(cot A+ 1 โ cosec A)

= (cot A โ cosec^{2}A + cot^{2}A + cosec A)/(cot A+ 1 โ cosec A) (using cosec^{2}A โ cot^{2}A = 1

= [(cot A + cosec A) โ (cosec^{2}A โ cot^{2}A)]/(cot A+ 1 โ cosec A)

= [(cot A + cosec A) โ (cosec A + cot A)(cosec A โ cot A)]/(1 โ cosec A + cot A)

= (cot A + cosec A)(1 โ cosec A + cot A)/(1 โ cosec A + cot A)

= cot A + cosec A = R.H.S.

Therefore, (cos Aโsin A+1)/(cos A+sin Aโ1) = cosec A + cot A

Hence Proved

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

We know that 1/cos A = sec A and sin A/ cos A = tan A and it becomes,

= โ(sec A+ tan A)/(sec A-tan A)

Now using rationalization, we get

= (sec A + tan A)/1

= sec A + tan A = R.H.S

Hence proved

(vii) (sin ฮธ โ 2sin^{3}ฮธ)/(2cos^{3}ฮธ-cos ฮธ) = tan ฮธ

L.H.S. = (sin ฮธ โ 2sin^{3}ฮธ)/(2cos^{3}ฮธ โ cos ฮธ)

Take sin ฮธ as in numerator and cos ฮธ in denominator as outside, it becomes

= [sin ฮธ(1 โ 2sin^{2}ฮธ)]/[cos ฮธ(2cos^{2}ฮธ- 1)]

We know that sin^{2}ฮธ = 1-cos^{2}ฮธ

= sin ฮธ[1 โ 2(1-cos^{2}ฮธ)]/[cos ฮธ(2cos^{2}ฮธ -1)]

= [sin ฮธ(2cos^{2}ฮธ -1)]/[cos ฮธ(2cos^{2}ฮธ -1)]

= tan ฮธ = R.H.S.

Hence proved

(viii) (sin A + cosec A)^{2 }+ (cos A + sec A)^{2} = 7+tan^{2}A+cot^{2}A

L.H.S. = (sin A + cosec A)^{2 }+ (cos A + sec A)^{2}

It is of the form (a+b)^{2}, expand it

(a+b)^{2} =a^{2} + b^{2} +2ab

^{ }= (sin^{2}A + cosec^{2}A + 2 sin A cosec A) + (cos^{2}A + sec^{2}A + 2 cos A sec A)

= (sin^{2}A + cos^{2}A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan^{2}A + 1 + cot^{2}A

= 1 + 2 + 2 + 2 + tan^{2}A + cot^{2}A

= 7+tan^{2}A+cot^{2}A = R.H.S.

Therefore, (sin A + cosec A)^{2 }+ (cos A + sec A)^{2} = 7+tan^{2}A+cot^{2}A

Hence proved.

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

First, find the simplified form of L.H.S

L.H.S. = (cosec A โ sin A)(sec A โ cos A)

Now, substitute the inverse and equivalent trigonometric ratio forms

= (1/sin A โ sin A)(1/cos A โ cos A)

= [(1-sin^{2}A)/sin A][(1-cos^{2}A)/cos A]

= (cos^{2}A/sin A)ร(sin^{2}A/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/[(sin^{2}A+cos^{2}A)/sin A cos A]

= cos A sin A

L.H.S. = R.H.S.

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

Hence proved

(x) (1+tan^{2}A/1+cot^{2}A) = (1-tan A/1-cot A)^{2} =^{ }tan^{2}A

L.H.S. = (1+tan^{2}A/1+cot^{2}A)

Since cot function is the inverse of tan function,

= (1+tan^{2}A/1+1/tan^{2}A)

= 1+tan^{2}A/[(1+tan^{2}A)/tan^{2}A]

Now cancel the 1+tan^{2}A terms, we get

= tan^{2}A

(1+tan^{2}A/1+cot^{2}A) = tan^{2}A

Similarly,

(1-tan A/1-cot A)^{2} =^{ }tan^{2}A

Hence proved

## NCERT Solutions for Class 10 Chapter 8 โ Introduction to Trigonometry

For the **Class 10 CBSE** Maths paper, out of the 80 marks, 12 marks are assigned from the unit 5 โTrigonometryโ. The paper consists of 4 parts. Each part carries different marks and the questions have been assigned with 1 mark, two marks, 3 marks and 4 marks. You can expect at least 2-3 compulsory questions from this chapter. The main topics covered in this chapter include:

### 8.1 Introduction

You have already studied about triangles, and in particular, right triangles, in your earlier classes. In this chapter, we will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle. We also define the trigonometric ratios for angles of measure 0o and 90o. We will calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities.

### 8.2 Trigonometric Identities

You have studied the concept of ratio, in your earlier classes. We now define certain ratios involving the sides of a right triangle, and call them trigonometric ratios. The topic is explained with suitable examples by using different functions of Trigonometry.

### 8.3 Trigonometric Ratios of Some Specific Angles

From Geometry, you are already familiar with the construction of angles of 300, 45o, 60o and 900. In this section, we will find the values of the trigonometric ratios for these angles and for 0o. It explains Trigonometric Ratios of 45o, Trigonometric Ratios of 30o and 600, Trigonometric Ratios of 0o and 90o with suitable examples.

### 8.4 Trigonometric Ratios of Complementary Angles

Two angles are said to be complementary if their sum equals 90o. The topic discusses various formulas to solve numerical problems related to trigonometric ratios.

### 8.5 Trigonometric Identities

You may recall that an equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true of all values of the angles involved. In this section, we will prove one trigonometric identity, and use it further to prove other useful trigonometric identities.

### 8.6 Summary

Summary is the brief of concepts that includes all the important points which you need to memorize to solve numerical problems related to the chapter.

**List of Exercises in Class 10 Maths Chapter 8 :**

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

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

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

Exercise 8.4 Solutions โ 5 Questions ( 2 short answers, 2 long answers, 1 MCQ)

Hence, these NCERT Solutions for Class 10 Maths will help students understand different types of questions and their answers along with key shortcuts and diagrammatic representations. All the **NCERT Solutions for Class 10 Maths Chapter 8** given here are presented in simple language. Comprehending these solutions thoroughly will aid students to solve complex problems effortlessly.

The faculty have curated the **NCERT Solutions for Class 10** in a lucid manner to improve the problem-solving abilities among the students. For a more clear idea about Introduction To Trigonometry, students can refer to the study materials available at BYJUโS.

Also Access |

NCERT Exemplar for Class 10 Maths Chapter 8 Introduction To Trigonometry |

CBSE Notes for Class 10 Maths Chapter 8 Introduction To Trigonometry |

## Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 8

### List out the frequently asked topics of chapter 8 of NCERT Solutions in the board exam of Class 10 Maths?

The frequently asked topics of Chapter 8 of NCERT Solutions in the board exam of Class 10 Maths are introduction to trigonometry, trigonometric identities, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles and trigonometric identities.

### Why should we download NCERT Solutions for Class 10 Maths Chapter 8 from BYJUโS?

BYJUโS provides the most accurate answers for the questions present in the NCERT Solutions for Class 10 Maths Chapter 8. These solutions can be viewed online as well as downloaded in PDF format. The solutions of this chapter are explained by experts very clearly with neat diagrams wherever necessary.

### Is NCERT Solutions for Class 10 Maths Chapter 8 important from the exam point of view?

Yes, all the chapters present in NCERT Solutions for Class 10 Maths are important for board exams as well as for higher grades. Students should practice all the questions present in NCERT Solutions for Class 10 Maths Chapter 8 to procure high marks.