NCERT Solutions for Class 10 Math Chapter 8 Introduction To Trigonometry are provided here with simple step-by-step explanations. These solutions for Introduction To Trigonometry are extremely popular among Class 10 students for Math Introduction To Trigonometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 10 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### 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

AC2 = AB2 + BC2

= (24 cm)2 + (7 cm)2

= (576 + 49) cm2

= 625 cm2

∴ 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 math - Introduction to Trigonometry 181 , Question 1

#### Question 2:

In the given figure find tan P − cot R

#### Answer:

Applying Pythagoras theorem for ΔPQR, we obtain

PR2 = PQ2 + QR2

(13 cm)2 = (12 cm)2 + QR2

169 cm2 = 144 cm2 + QR2

25 cm2 = QR2

QR = 5 cm

tan P − cot R =

##### Video Solution for Introduction to Trigonometry (Page: 181 , Q.No.: 2)

NCERT Solution for Class 10 math - Introduction to Trigonometry 181 , Question 2

#### 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 3k. Therefore, AC will be 4k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC2 = AB2 + BC2

(4k)2 = AB2 + (3k)2

16k 2 − 9k 2 = AB2

7k 2 = AB2

AB =

##### Video Solution for Introduction to Trigonometry (Page: 181 , Q.No.: 3)

NCERT Solution for Class 10 math - Introduction to Trigonometry 181 , Question 3

#### 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 8k.Therefore, BC will be 15k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC2 = AB2 + BC2

= (8k)2 + (15k)2

= 64k2 + 225k2

= 289k2

AC = 17k

##### Video Solution for Introduction to Trigonometry (Page: 181 , Q.No.: 4)

NCERT Solution for Class 10 math - Introduction to Trigonometry 181 , Question 4

#### 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 13k, AB will be 12k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

(AC)2 = (AB)2 + (BC)2

(13k)2 = (12k)2 + (BC)2

169k2 = 144k2 + BC2

25k2 = BC2

BC = 5k

##### Video Solution for Introduction to Trigonometry (Page: 181 , Q.No.: 5)

NCERT Solution for Class 10 math - Introduction to Trigonometry 181 , Question 5

#### 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

CD2 = AC2 − AD2 … (3)

And, CD2 = BC2 − BD2 … (4)

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

AC2 − AD2 = BC2 − BD2

⇒ (k BC)2 − (k BD)2 = BC2 − BD2

k2 (BC2 − BD2) = BC2 − BD2

k2 = 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 math - Introduction to Trigonometry 181 , Question 6

#### Question 7:

If cot θ =, evaluate

(i) (ii) cot2 θ

#### Answer:

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

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

Applying Pythagoras theorem in ΔABC, we obtain

AC2 = AB2 + BC2

= (8k)2 + (7k)2

= 64k2 + 49k2

= 113k2

AC =

(i)

(ii) cot2 θ = (cot θ)2 = =

##### Video Solution for Introduction to Trigonometry (Page: 181 , Q.No.: 7)

NCERT Solution for Class 10 math - Introduction to Trigonometry 181 , Question 7

#### 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 4k, then BC will be 3k, where k is a positive integer.

In ΔABC,

(AC)2 = (AB)2 + (BC)2

= (4k)2 + (3k)2

= 16k2 + 9k2

= 25k2

AC = 5k

cos2 A − sin2 A =

##### Video Solution for Introduction to Trigonometry (Page: 181 , Q.No.: 8)

NCERT Solution for Class 10 math - Introduction to Trigonometry 181 , Question 8

#### 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,

AC2 = AB2 + BC2

=

= 3k2 + k2 = 4k2

∴ AC = 2k

(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 math - Introduction to Trigonometry 181 , Question 9

#### 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

PR2 = PQ2 + QR2

x2 = (5)2 + (25 − x)2

x2 = 25 + 625 + x2 − 50x

50x = 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 math - Introduction to Trigonometry 181 , Question 10

#### 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 12k, AB will be 5k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC2 = AB2 + BC2

(12k)2 = (5k)2 + BC2

144k2 = 25k2 + BC2

BC2 = 119k2

BC = 10.9k

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

BC should be such that,

AC − AB < BC < AC + AB

12k − 5k < BC < 12k + 5k

7k < BC < 17 k

However, BC = 10.9k. 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

#### Question 1:

Evaluate the following

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan245° + cos230° − sin260°

(iii)

(iv)

(v)

#### Answer:

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan245° + cos230° − sin260°

(iii)

(iv)

(v)

##### Video Solution for Introduction to Trigonometry (Page: 187 , Q.No.: 1)

NCERT Solution for Class 10 math - Introduction to Trigonometry 187 , Question 1

#### 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 math - Introduction to Trigonometry 187 , Question 2

#### 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 math - Introduction to Trigonometry 187 , Question 3

#### 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 math - Introduction to Trigonometry 187 , Question 4

#### 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 math - Introduction to Trigonometry 189 , Question 1

#### 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 math - Introduction to Trigonometry 189 , Question 2

#### 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 math - Introduction to Trigonometry 189 , Question 3

#### 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 math - Introduction to Trigonometry 189 , Question 4

#### 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 math - Introduction to Trigonometry 189 , Question 5

#### 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 math - Introduction to Trigonometry 190 , Question 6

#### 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 math - Introduction to Trigonometry 190 , Question 7

#### 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 math - Introduction to Trigonometry 193 , Question 1

#### Question 2:

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

#### Answer:

We know that,

Also, sin2 A + cos2 A = 1

sin2 A = 1 − cos2 A

tan2A + 1 = sec2A

tan2A = sec2A − 1

##### Video Solution for Introduction to Trigonometry (Page: 193 , Q.No.: 2)

NCERT Solution for Class 10 math - Introduction to Trigonometry 193 , Question 2

#### Question 3:

Evaluate

(i)

(ii) sin25° cos65° + cos25° sin65°

#### Answer:

(i)

(As sin2A + cos2A = 1)

= 1

(ii) sin25° cos65° + cos25° sin65°

= sin225° + cos225°

= 1 (As sin2A + cos2A = 1)

##### Video Solution for Introduction to Trigonometry (Page: 193 , Q.No.: 3)

NCERT Solution for Class 10 math - Introduction to Trigonometry 193 , Question 3

#### Question 4:

Choose the correct option. Justify your choice.

(i) 9 sec2 A − 9 tan2 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) sec2 A

(B) −1

(C) cot2 A

(D) tan2 A

#### Answer:

(i) 9 sec2A − 9 tan2A

= 9 (sec2A − tan2A)

= 9 (1) [As sec2 A − tan2 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 math - Introduction to Trigonometry 193 , Question 4

#### 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 cosec2 = 1 + cot2,

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 math - Introduction to Trigonometry 193 , Question 5

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