**Introduction to Trigonometry:**In the beginning, a quote is given about trigonometry in an eye-catching way. Some real-life examples like finding the height of Qutub Minar, finding the width of the river, and finding the altitude of the ground from the hot air balloon are given in order to explain the need for

**trigonometry**. In this chapter, students will study the

**trigonometric ratios**of the

**angle**i.e

**ratios of the sides of a right triangle with respect to its acute angles**. The discussion of the

**trigonometric**

**ratios**will be restricted to

**acute angles**only.

In section 8.2 various

**trigonometric ratios**are explained.

**Trigonometric ratios**are given in this section:

**SINE****COSINE****TANGENT****COSECANT****SECANT****COTANGENT**

**trigonometric ratios**and how to prove certain given relations.

- Special note is given stating that the values of the
**trigonometric ratios**of an**angle**do not vary with the lengths of the sides of the**triangle**if the**angle**remains the same.

**trigonometric ratios**.

In the next section,

**trigonometric ratios of some specific angles**are given. Value of ratios of

**angle 0$\xb0$**,

**30$\xb0$, 45**$\xb0$,

**60$\xb0$**,

**90$\xb0$**are given in table 8.1. Students must learn these values as they are the most important part of this chapter and are essential to solve the problems not only in this chapter but are useful in higher grades as well.

Exercise 8.2 contains 4 different types of questions based on

**trigonometric ratios.**

In section 8.4 t

**rigonometric ratios**of

**complementary angles**are given. Total 6 relations are given in this section. These relations must be on the fingertips of students in order to score better. Exercise 8.3 is based on the same.

The next part of the chapter is about

**trigonometric identities**. Three important

**identities**are highlighted in this section. These identities are true for all

**angles**(

*A*) such that

**. Using the same students will be able to solve the problems of exercise 8.4 including the problems which need certain given statements to prove. In the end key points of chapter are discussed.**

*A*$\in $[0$\xb0$, 90$\xb0$]#### Page No 181:

#### Question 1:

In ΔABC right angled at B, AB = 24 cm, BC = 7 cm. Determine

(i) sin A, cos A

(ii) sin C, cos C

#### Answer:

Applying Pythagoras theorem for ΔABC, we obtain

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

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

= (576 + 49) cm^{2}

= 625 cm^{2}

∴ AC = cm = 25 cm

(i) sin A =

cos A =

(ii)

sin C =

cos C =

##### Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 1)

NCERT Solution for Class 10 maths - introduction to trigonometry 181 , Question 1

#### Page No 181:

#### Question 2:

In the given figure find tan P − cot R

#### Answer:

Applying Pythagoras theorem for ΔPQR, we obtain

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

(13 cm)^{2} = (12 cm)^{2} + QR^{2}

169 cm^{2} = 144 cm^{2} + QR^{2}

25 cm^{2} = QR^{2}

QR = 5 cm

tan P − cot R =

##### Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 2)

NCERT Solution for Class 10 maths - introduction to trigonometry 181 , Question 2

#### Page No 181:

#### Question 3:

If sin A =, calculate cos A and tan A.

#### Answer:

Let ΔABC be a right-angled triangle, right-angled at point B.

Given that,

Let BC be 3*k*. Therefore, AC will be 4*k*, where *k* is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

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

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

16*k* ^{2} − 9*k* ^{2} = AB^{2}

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

AB =

##### Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 3)

NCERT Solution for Class 10 maths - introduction to trigonometry 181 , Question 3

#### Page No 181:

#### Question 4:

Given 15 cot A = 8. Find sin A and sec A

#### Answer:

Consider a right-angled triangle, right-angled at B.

It is given that,

cot A =

Let AB be 8*k*.Therefore, BC will be 15*k*, where *k* is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

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

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

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

= 289*k*^{2}

AC = 17*k*

##### Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 4)

NCERT Solution for Class 10 maths - introduction to trigonometry 181 , Question 4

#### Page No 181:

#### Question 5:

Given sec θ =, calculate all other trigonometric ratios.

#### Answer:

Consider a right-angle triangle ΔABC, right-angled at point B.

If AC is 13*k*, AB will be 12*k*, where *k* is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

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

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

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

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

BC = 5*k*

##### Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 5)

NCERT Solution for Class 10 maths - introduction to trigonometry 181 , Question 5

#### Page No 181:

#### Question 6:

If ∠A and ∠B are acute angles such that cos A = cos B, then show that

∠A = ∠B.

#### Answer:

Let us consider a triangle ABC in which CD ⊥ AB.

It is given that

cos A = cos B

… (1)

We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP.

From equation (1), we obtain

By using the converse of B.P.T,

CD||BP

⇒∠ACD = ∠CPB (Corresponding angles) … (3)

And, ∠BCD = ∠CBP (Alternate interior angles) … (4)

By construction, we have BC = CP.

∴ ∠CBP** = **∠CPB (Angle opposite to equal sides of a triangle) … (5)

From equations (3), (4), and (5), we obtain

∠ACD = ∠BCD … (6)

In ΔCAD and ΔCBD,

∠ACD = ∠BCD [Using equation (6)]

∠CDA = ∠CDB [Both 90°]

Therefore, the remaining angles should be equal.

∴∠CAD = ∠CBD

⇒ ∠A = ∠B

**Alternatively,**

Let us consider a triangle ABC in which CD ⊥ AB.

It is given that,

cos A = cos B

Let

⇒ AD = *k* BD … (1)

And, AC = *k* BC … (2)

Using Pythagoras theorem for triangles CAD and CBD, we obtain

CD^{2} = AC^{2} − AD^{2} … (3)

And, CD^{2} = BC^{2} − BD^{2} … (4)

From equations (3) and (4), we obtain

AC^{2} − AD^{2} = BC^{2} − BD^{2}

⇒ (*k *BC)^{2} − (*k* BD)^{2} = BC^{2} − BD^{2}

⇒ *k*^{2} (BC^{2} − BD^{2}) = BC^{2} − BD^{2}

⇒ *k*^{2} = 1

⇒ *k* = 1

Putting this value in equation (2), we obtain

AC = BC

⇒ ∠A = ∠B(Angles opposite to equal sides of a triangle)

##### Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 6)

NCERT Solution for Class 10 maths - introduction to trigonometry 181 , Question 6

#### Page No 181:

#### Question 7:

If cot θ =, evaluate

(i) (ii) cot^{2} θ

#### Answer:

Let us consider a right triangle ABC, right-angled at point B.

If BC is 7*k*, then AB will be 8*k*, where *k* is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

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

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

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

= 113*k*^{2}

AC =

(i)

(ii) cot^{2} θ = (cot θ)^{2} = =

##### Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 7)

NCERT Solution for Class 10 maths - introduction to trigonometry 181 , Question 7

#### Page No 181:

#### Question 8:

If 3 cot A = 4, Check whether

#### Answer:

It is given that 3cot A = 4

Or, cot A =

Consider a right triangle ABC, right-angled at point B.

If AB is 4*k*, then BC will be 3*k*, where *k* is a positive integer.

In ΔABC,

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

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

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

= 25*k*^{2}

AC = 5*k*

cos^{2} A − sin^{2} A =

∴

##### Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 8)

NCERT Solution for Class 10 maths - introduction to trigonometry 181 , Question 8

#### Page No 181:

#### Question 9:

In ΔABC, right angled at B. If, find the value of

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

(ii) cos A cos C − sin A sin C

#### Answer:

If BC is *k*, then AB will be, where* k* is a positive integer.

In ΔABC,

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

=

= 3*k*^{2} + *k*^{2} = 4*k*^{2}

∴ AC = 2*k*

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

(ii) cos A cos C − sin A sin C

##### Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 9)

NCERT Solution for Class 10 maths - introduction to trigonometry 181 , Question 9

#### Page No 181:

#### Question 10:

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

#### Answer:

Given that, PR + QR = 25

PQ = 5

Let PR be *x*.

Therefore, QR = 25 − *x*

Applying Pythagoras theorem in ΔPQR, we obtain

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

*x*^{2} = (5)^{2} + (25 − *x*)^{2}

*x*^{2} = 25 + 625 + *x*^{2} − 50*x*

50*x* = 650

*x* = 13

Therefore, PR = 13 cm

QR = (25 − 13) cm = 12 cm

##### Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 10)

NCERT Solution for Class 10 maths - introduction to trigonometry 181 , Question 10

#### Page No 181:

#### Question 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 =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 θ =, for some angle θ

#### Answer:

(i) Consider a ΔABC, right-angled at B.

But > 1

∴tan A > 1

So, tan A < 1 is not always true.

Hence, the given statement is false.

(ii)

Let AC be 12*k*, AB will be 5*k*, where *k* is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

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

(12*k*)^{2} = (5*k*)^{2} + BC^{2}

144*k*^{2} = 25*k*^{2} + BC^{2}

BC^{2} = 119*k*^{2}

BC = 10.9*k*

It can be observed that for given two sides AC = 12*k* and AB = 5*k*,

BC should be such that,

AC − AB < BC < AC + AB

12*k* − 5*k* < BC < 12*k* + 5*k*

7*k *< BC < 17 *k*

However, BC = 10.9*k*. Clearly, such a triangle is possible and hence, such value of sec A is possible.

Hence, the given statement is true.

(iii) Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A.

Hence, the given statement is false.

(iv) cot A is not the product of cot and A. It is the cotangent of ∠A.

Hence, the given statement is false.

(v) sin θ =

We know that in a right-angled triangle,

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible.

Hence, the given statement is false

##### Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 11)

NCERT Solution for Class 10 maths - introduction to trigonometry 181 , Question 11

#### Page No 187:

#### Question 1:

Evaluate the following

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan^{2}45° + cos^{2}30° − sin^{2}60°

(iii)

(iv)

(v)

#### Answer:

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan^{2}45° + cos^{2}30° − sin^{2}60°

(iii)

(iv)

(v)

##### Video Solution for introduction to trigonometry (Page: 187 , Q.No.: 1)

NCERT Solution for Class 10 maths - introduction to trigonometry 187 , Question 1

#### Page No 187:

#### Question 2:

Choose the correct option and justify your choice.

(i)

(A). sin60°

(B). cos60°

(C). tan60°

(D). sin30°

(ii)

(A). tan90°

(B). 1

(C). sin45°

(D). 0

(iii) sin2A = 2sinA is true when A =

(A). 0°

(B). 30°

(C). 45°

(D). 60°

(iv)

(A). cos60°

(B). sin60°

(C). tan60°

(D). sin30°

#### Answer:

(i)

Out of the given alternatives, only

Hence, (A) is correct.

(ii)

Hence, (D) is correct.

(iii)Out of the given alternatives, only A = 0° is correct.

As sin 2A = sin 0° = 0

2 sinA = 2sin 0° = 2(0) = 0

Hence, (A) is correct.

(iv)

Out of the given alternatives, only tan 60°

Hence, (C) is correct.

##### Video Solution for introduction to trigonometry (Page: 187 , Q.No.: 2)

NCERT Solution for Class 10 maths - introduction to trigonometry 187 , Question 2

#### Page No 187:

#### Question 3:

If and;

0° < A + B ≤ 90°, A > B find A and B.

#### Answer:

⇒

⇒ A + B = 60 … (1)

⇒ tan (A − B) = tan30

⇒ A − B = 30 … (2)

On adding both equations, we obtain

2A = 90

⇒ A = 45

From equation (1), we obtain

45 + B = 60

B = 15

Therefore, ∠A = 45° and ∠B = 15°

##### Video Solution for introduction to trigonometry (Page: 187 , Q.No.: 3)

NCERT Solution for Class 10 maths - introduction to trigonometry 187 , Question 3

#### Page No 187:

#### Question 4:

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

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

(ii) The value of sinθ increases as θ increases

(iii) The value of cos θ increases as θ increases

(iv) sinθ = cos θ for all values of θ

(v) cot A is not defined for A = 0°

#### Answer:

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

Let A = 30° and B = 60°

sin (A + B) = sin (30° + 60°)

= sin 90°

= 1

sin A + sin B = sin 30° + sin 60°

Clearly, sin (A + B) ≠ sin A + sin B

Hence, the given statement is false.

(ii) The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as

sin 0° = 0

sin 90° = 1

Hence, the given statement is true.

(iii) cos 0° = 1

cos90° = 0

It can be observed that the value of cos θ does not increase in the interval of 0° < θ < 90°.

Hence, the given statement is false.

(iv) sin θ = cos θ for all values of θ.

This is true when θ = 45°

As

It is not true for all other values of θ.

As and ,

Hence, the given statement is false.

(v) cot A is not defined for A = 0°

As ,

= undefined

Hence, the given statement is true.

##### Video Solution for introduction to trigonometry (Page: 187 , Q.No.: 4)

NCERT Solution for Class 10 maths - introduction to trigonometry 187 , Question 4

#### Page No 189:

#### Question 1:

Evaluate

(I)

(II)

(III) cos 48° − sin 42°

(IV)cosec 31° − sec 59°

#### Answer:

(I)

(II)

(III)cos 48° − sin 42° = cos (90°− 42°) − sin 42°

= sin 42° − sin 42°

= 0

(IV) cosec 31° − sec 59° = cosec (90° − 59°) − sec 59°

= sec 59° − sec 59°

= 0

##### Video Solution for introduction to trigonometry (Page: 189 , Q.No.: 1)

NCERT Solution for Class 10 maths - introduction to trigonometry 189 , Question 1

#### Page No 189:

#### Question 2:

Show that

(I) tan 48° tan 23° tan 42° tan 67° = 1

(II)cos 38° cos 52° − sin 38° sin 52° = 0

#### Answer:

(I) tan 48° tan 23° tan 42° tan 67°

= tan (90° − 42°) tan (90° − 67°) tan 42° tan 67°

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°)

= (1) (1)

= 1

(II) cos 38° cos 52° − sin 38° sin 52°

= cos (90° − 52°) cos (90°−38°) − sin 38° sin 52°

= sin 52° sin 38° − sin 38° sin 52°

= 0

##### Video Solution for introduction to trigonometry (Page: 189 , Q.No.: 2)

NCERT Solution for Class 10 maths - introduction to trigonometry 189 , Question 2

#### Page No 189:

#### Question 3:

If tan 2A = cot (A− 18°), where 2A is an acute angle, find the value of A.

#### Answer:

Given that,

tan 2A = cot (A− 18°)

cot (90° − 2A) = cot (A −18°)

90° − 2A = A− 18°

108° = 3A

A = 36°

##### Video Solution for introduction to trigonometry (Page: 189 , Q.No.: 3)

NCERT Solution for Class 10 maths - introduction to trigonometry 189 , Question 3

#### Page No 189:

#### Question 4:

If tan A = cot B, prove that A + B = 90°

#### Answer:

Given that,

tan A = cot B

tan A = tan (90° − B)

A = 90° − B

A + B = 90°

##### Video Solution for introduction to trigonometry (Page: 189 , Q.No.: 4)

NCERT Solution for Class 10 maths - introduction to trigonometry 189 , Question 4

#### Page No 189:

#### Question 5:

If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.

#### Answer:

Given that,

sec 4A = cosec (A − 20°)

cosec (90° − 4A) = cosec (A − 20°)

90° − 4A= A− 20°

110° = 5A

A = 22°

##### Video Solution for introduction to trigonometry (Page: 189 , Q.No.: 5)

NCERT Solution for Class 10 maths - introduction to trigonometry 189 , Question 5

#### Page No 190:

#### Question 6:

If A, Band C are interior angles of a triangle ABC then show that

#### Answer:

We know that for a triangle ABC,

∠ A + ∠B + ∠C = 180°

∠B + ∠C= 180° − ∠A

##### Video Solution for introduction to trigonometry (Page: 190 , Q.No.: 6)

NCERT Solution for Class 10 maths - introduction to trigonometry 190 , Question 6

#### Page No 190:

#### Question 7:

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

#### Answer:

sin 67° + cos 75°

**= **sin (90° − 23°) + cos (90° − 15°)

**= **cos 23° + sin 15°

##### Video Solution for introduction to trigonometry (Page: 190 , Q.No.: 7)

NCERT Solution for Class 10 maths - introduction to trigonometry 190 , Question 7

#### Page No 193:

#### Question 1:

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

#### Answer:

We know that,

will always be positive as we are adding two positive quantities.

Therefore,

We know that,

However,

Therefore,

Also,

##### Video Solution for introduction to trigonometry (Page: 193 , Q.No.: 1)

NCERT Solution for Class 10 maths - introduction to trigonometry 193 , Question 1

#### Page No 193:

#### Question 2:

Write all the other trigonometric ratios of ∠A in terms of sec A.

#### Answer:

We know that,

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

sin^{2} A = 1 − cos^{2} A

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

tan^{2}A = sec^{2}A − 1

##### Video Solution for introduction to trigonometry (Page: 193 , Q.No.: 2)

NCERT Solution for Class 10 maths - introduction to trigonometry 193 , Question 2

#### Page No 193:

#### Question 3:

Evaluate

(i)

(ii) sin25° cos65° + cos25° sin65°

#### Answer:

(i)

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

= 1

(ii) sin25° cos65° + cos25° sin65°

= sin^{2}25° + cos^{2}25°

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

##### Video Solution for introduction to trigonometry (Page: 193 , Q.No.: 3)

NCERT Solution for Class 10 maths - introduction to trigonometry 193 , Question 3

#### Page No 193:

#### Question 4:

Choose the correct option. Justify your choice.

(i) 9 sec^{2} A − 9 tan^{2} A =

(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) (secA + tanA) (1 − sinA) =

(A) secA

(B) sinA

(C) cosecA

(D) cosA

(iv)

(A) sec^{2 }A

(B) −1

(C) cot^{2 }A

(D) tan^{2 }A

#### Answer:

(i) 9 sec^{2}A − 9 tan^{2}A

= 9 (sec^{2}A − tan^{2}A__)__

= 9 (1) [As sec^{2} A − tan^{2} A = 1]

= 9

Hence, alternative (B) is correct.

(ii)

(1 + tan θ + sec θ) (1 + cot θ − cosec θ)

Hence, alternative (C) is correct.

(iii) (secA + tanA) (1 − sinA)

= cosA

Hence, alternative (D) is correct.

(iv)

Hence, alternative (D) is correct.

##### Video Solution for introduction to trigonometry (Page: 193 , Q.No.: 4)

NCERT Solution for Class 10 maths - introduction to trigonometry 193 , Question 4

#### Page No 193:

#### Question 5:

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

#### Answer:

(i)

(ii)

(iii)

= secθ cosec θ +

= R.H.S.

(iv)

= R.H.S

(v)

Using the identity cosec^{2} = 1 + cot^{2},

L.H.S =

= cosec A + cot A

= R.H.S

(vi)

(vii)

(viii)

(ix)

Hence, L.H.S = R.H.S

(x)

##### Video Solution for introduction to trigonometry (Page: 193 , Q.No.: 5)

NCERT Solution for Class 10 maths - introduction to trigonometry 193 , Question 5

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