NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 8
Chapter Name : Introduction to Trigonometry

Exercise 8.1

Q1 In △ ABC right angled at B, AB=24 cm, BC=7 m. Determine
(i) sin A, cos A
(ii) sin C, cos C

Applying Pythagoras theorem for ΔABC, we obtain

2 2 2
AC = AB + BC

2 2
= (24cm) + (7cm)
Answer. 2
= (576 + 49)cm

2
= 625cm

∴ AC = √625cm = 25cm

Side opposite to ∠A BC
sin A = =
Hypotenuse AC

7
=
25

Side adjacent to ∠A AB 24
cos A = = =
Hypotenuse AC 25

(ii)
Side opposite to ∠C AB
sin C = =
Hypotenuse AC

24
=
25

Side adjacent to ∠C BC
cos C = =
Hypotenuse AC

7
=
25

Page : 181, Block Name : Exercise 8.1

Q2 In the given gure nd tan P - cot R.

Applying Pythagoras theorem for ΔP QR, we obtain

2 2 2
PR = PQ + QR

2 2 2
(13cm) = (12cm) + QR
Answer. 2 2 2
169cm = 144cm + QR

2 2
25cm = QR

QR = 5cm

Side opposite to ∠P QR
tan P = =
Side adjacent to ∠P PQ

5
=
12

Side adjacent to ∠R QR
cot R = =
Side opposite to ∠R PQ

5
=
12
5 5
tan P − cot R = − = 0
12 12

Page : 181 , Block Name : Exercise 8.1

Q3 If sin A =
3
, calculate cos A and tan A.
4

Answer.LetΔABC be a right-angled triangle, right-angled at point B

Given that,
3
sin A =
4

BC 3
=
AC 4

Let BC be 3k . Therefore, AC will be 4k , where k is a positive integer.
Applying Pythagoras theorem in △ABC , we obtain
2 2 2
AC = AB + BC

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

2 2 2
16k − 9k = AB

2 2
7k = AB

AB = √7k

Side adjacent to ∠A
cos A =
Hypotent t0∠A

AB √7k √7
= = =
AC 4k 4

Side opposite to ∠A
tan A =
Side adjacent to ∠A

BC 3k 3
= = =
AB √7k √7

Page : 181 , Block Name : Exercise 8.1

Q4 Given 15 cot A = 8. Find sin A and sec A.

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

Given that,
3
sin A =
4

BC 3
=
AC 4

Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer.
Applying Pythagoras theorem in △ABC , we obtain
2 2 2
AC = AB + BC

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

2 2 2
16k − gk = AB

2 2
7k = AB

AB = √7k

Side adjacent to ∠A
cos A =
Hypotenuse

AB √7k √7
= = =
AC 4k 4
Side opposite to ∠A
tan A =
Side adjacent to ∠A

BC 3k 3
= = =
AB √7k √7

Side adjacent to ∠A
cot A =
Side opposite to ∠A

AB
=
BC
It is given that,

8
cot A =
15

AB 8
=
BC 15

Let AB be 8k.Therefore, 3C will be 15k, where k is a positive integer.
Applying Pythagoras theorem in △ABC, we obtain
2 2 2
AC = AB + BC

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

2 2
= 64k + 225k

2
= 289k

AC = 17k

Side opposite to ∠A BC
sin A = =
Hypotenuse AC

15k 15
= =
17k 17

Hypotenuse
sec A =
Side adjacent to ∠A

AC 17
= =
AB 8

Page : 181 , Block Name : Exercise 8.1

Q5 Given sec θ = calculate all other trigonometric ratios.
13
,
12

Answer. Consider a right-angle △ABC, right-angled at point B

Hypotenuse
sec θ =
Side adjacent to ∠θ

13 AC
=
12 AB

If AC is 13k, AB will be 12k, where k is a positive integer.
Applying Pythagoras theorem in AABC, we obtain

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

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

2 2 2
169k = 144k + BC

2 2
25k = BC

BC = 5k
Side opposite to ∠θ BC 5k 5
sin θ = = = =
Hypotenuse AC 13k 13

Side adjacent to ∠θ AB 12k 12
cos θ = = = =
Hypotenuse AC 13k 13

Side opposite to ∠θ BC 5k 5
tan θ = = = =
Side adjacent to ∠θ AB 12k 12

Side adjacent to ∠θ AB 12k 12
cot θ = = = =
Side opposite to ∠θ BC 5k 5

Hypotenuse AC 13k 13
cosec θ = = = =
Side opposite to ∠θ BC 5k 5

Page : 181 , Block Name : Exercise 8.1

If ∠A and ∠B are acute angles such that cos A = cos B , then show that
Q6
∠A = ∠B

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

It is given that,
cos A = cos B

AD BD
⇒ = …
AC BC

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

From equation (1) we obtain,

AD AC
=
BD BC
( By construction, we have BC = CP)
AD AC
⇒ =
BD CP

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)
∘
∠CDA = ∠CDB [Both90 ]

Therefore, the remaining angles should be equal.

∴ ∠CAD = ∠CBD

⇒ ∠A = ∠B

Alternatively,
Let us consider triangle ABC in which CD ⊥ AB

It is given that,
Cos A = Cos B
AD BD
⇒ =
AC BC

AD AC
⇒ =
BD BC

AD AC
Let = = k
BD BC

⇒ AD = kBD … (1)

And, AC = kBC … (2)

using Pythagoras theorem for triangles CAD and CBD, we obtain
2 2 2
CD = AC − AD … (3)

2 2 2
And, CD = BC − BD … (4)

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

2 2 2 2
AC − AD = BC − BD
2 2 2 2
⇒ (kBC) − (kBD) = BC − BD

2 2 2 2 2
⇒ k (BC − BD ) = BC − BD

2
⇒ k = 1

⇒ k = 1

Putting this value in equation (2), we obtain
AC = BC
⇒ ∠A = ∠B(Angles opposite to equal sides of a angle)

Page : 181 , Block Name : Exercise 8.1

7
If cot θ = , evaluate
8

Q7 (1+sin θ)(1−sin θ)
2
(i) ( (ii) cot θ
(1+cos θ)(1−cos θ)

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

Side adjacent to ∠θ BC
cot θ = =
Side opposite to ∠θ AB

7
=
8

If BC is 7k, then AB will be 8k, where k is a positive integer.
Applying Pythagoras theorem in △ABC, we obtain
2 2 2
AC = AB + BC

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

2 2
= 64k + 49k

2
= 113k

AC = √113k

Side opposite to ∠θ AB
sin θ = =
Hypotenuse AC

8k 8
= =
√113k √113

Side adjacent to ∠θ BC
cos θ = =
Hypotenuse AC

7k 7
= =
√113k √113
2
(1+sin θ)(1−sin θ) (1−sin θ)
(i) = 2
(1+cos θ)(1−cos θ) (1−cos θ)
2
8
64
1−( ) 1−
√113 113
= 2
=
49
7 1−
1−( ) 113
√113

49

113 49
= =
64 64
113

2

(ii) cot θ = (cot θ)
2 2 7 49
= ( ) =
8 64

Page : 181 , Block Name : Exercise 8.1

2

Q8 If 3 cot A = 4, Check whether or not.
1−tan A 2 2
2
= cos A − sin A
1+tan A

Answer. It is given that 3 Cot A = 4
Or, Cot A = 4

3

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

Side adjacent to ∠A
cot A =
Side opposite to ∠A

AB 4
=
BC 3

If AB is 4k , then BC will be 3k, where k is a positive integer.
In △ABC,
2 2 2
(AC) = (AB) + (BC)

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

2 2
= 16k + 9k

2
= 25k

AC = 5k

Side adjacent to ∠A AB
cos A = =
Hypotenuse AC

4k 4
= =
5k 5
Side opposite to ∠A BC
sin A = =
Hypotenuse AC

3k 3
= =
5k 5
Side opposite to ∠A BC
tan A = =
Hypotenuse AB

3k 3
= =
4k 4

2
3
9
2
1−( ) 1−
1−tan A 4 16

2
= 2
=
9
1+tan A 3 1+
1+( ) 16
4

7

16 7
= =
25 25
16

2 2
2 2 4 3
cos A − sin A = ( ) − ( )
5 5

16 9 7
= − =
25 25 25

2
1−tan A 2 2
∴ 2
= cos A − sin A
1+tan A

Page : 181 , Block Name : Exercise 8.1

1
In △ABC, right angled at B . If A = , find the value of
√3

Q9 (i) sin A cos C + cos A sin C

(ii) cos A cos C − sin A sin C

Answer.

1
tan A =
√3

BC 1
=
AB √3

If BC is k , then AB will be , where k is a positive integer.
In △ABC,
2 2 2
AC = AB + BC

2 2
= (√3k) + (k)

2 2 2
= 3k + k = 4k

∴ AC = 2k
Side opposite to ∠A BC k 1
sin A = = = =
Hypotenuse AC 2k 2

Side adjacent to ∠A AB √3k √3
cos A = = = =
Hypotenuse AC 2k 2

Side opposite to ∠C AB √3k √3
sin C = = = =
Hypotenuse AC 2k 2

Side adjacent to ∠C BC k 1
cos C = = = =
Hypotenuse AC 2k 2

(i) sin A cos C + cos A sin C
1 1 √3 √3 1 3
= ( )( ) + ( )( ) = +
2 2 2 2 4 4

4
= = 1
4

(ii) cos A cos C - sin A sin C
√3 1 1 √3 √3 √3
= ( )( ) − ( )( ) = − = 0
2 2 2 2 4 4

Page : 181 , Block Name : Exercise 8.1

Q10
In ΔPQR, right angled at Q, PR + QR = 25cm and PQ = 5cm. 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
2 2 2
PR = PQ + QR

2 2 2
x = (5) + (25 − x)

2 2
x = 25 + 625 + x − 50x

50x = 650

x = 13

Therefore, PR = 13cm

QR = (25 − 13)cm = 12cm

Side opposite to ∠P QR 12
sin P = = =
Hypotenuse PR 13

Side adjacent to ∠P PQ 5
cos P = = =
Hypotenuse PR 13

Side opposite to ∠P QR 12
tan P = = =
Side adjacent to ∠P PQ 5

Page : 181 , Block Name : Exercise 8.1

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

(1) The value of tan A is always less than 1 .
Q11 12
(ii) sec A = for some value of angle A.
5

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

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

4
(v) sin θ = , for some angle θ
3

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

Side opposite to ∠A
tan A =
Side adjacent to ∠A

12
=
5
12
But > 1
5

∴ tan A > 1

So, tan A < 1 is not always true.

Hence, the given statement is false.

(ii)
12
sec A =
5

Hypotenuse 12
=
Side adjacent to ∠A 5

AC 12
=
AB 5

Let AC be 12k, AB will be 5k, where k is a positive integer.
Applying Pythagoras theorem in Δ ABC, we obtain
2 2 2
AC = AB + BC

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

2 2 2
144k = 25k + BC

2 2
BC = 119k

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 < 17k
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 not A. It is the cotangent of ∠A.
Hence, the given statement is false.
(v) sin θ =
4

3

We know that in a right-angled triangle,
Side opposite to ∠θ
sin θ =
Hypotenuse

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.

Page : 181 , Block Name : Exercise 8.1

Exercise 8.2

Q1 Evaluate the following
(i) sin 60 cos 30 + sin 30 cos 60
∘ ∘ ∘ ∘

(ii) 2 tan 45 + cos 30 − sin 60
2 ∘ 2 ∘ 2 ∘

∘

(iii) cos 45
∘
sec 30 +csc 30
∘

∘ ∘ ∘

(iv)
sin 30 +tan 45 −csc 60
∘ ∘ ∘
sec 30 +cos 60 +cot 45
2 ∘ 2 ∘ 2 ∘

(v)
5 cos 60 +4 sec 30 −tan 45

2 ∘ ∘
sin 30 +cos2 30

Answer. (i) sin 60 cos 30 ∘ ∘
+ sin 30
∘ ∘
cos 60
√3 √3 1 1
= ( )( ) + ( )( )
2 2 2 2

3 1 4
= + = = 1
4 4 4

(ii) 2 tan 45 2 ∘
+ cos
2
30
∘
− sin
2 ∘
60
2 2
√3 √3
2
= 2(1) + ( ) − ( )
2 2

3 3
= 2 + − = 2
4 4
∘

(iii) cos 45
∘
sec 30 +cosec30
∘

1 1

√2 √2

= =
2 2+2√3
+2
√3
√3

√3 √3
= =
√2(2+2√3) 2√2+2√6

√3(2√6−2√2)
=
(2√6+2√2)(2√6−2√2)

2√3(√6−√2) 2√3(√6−√2) 2√3(√6−√2)
= = =
(2√6)2 −(2√2)2 24−8 16

√18−√6 3√2−√6
= =
8 8
∘ ∘ ∘

(iv)
sin 30 +tan 45 −cosec60
∘ ∘ ∘
sec 30 +cos 60 +cot 45
1 2 3 2
+1− −
2 √3 2 √3

= =
2 1 3 2
+ +1 +
√3 2 2 √3

3√3−4 (3√3−4)
= =
3√3−4 (3√3+4)
2√3+4

2
(3√3−4)(3√3−4) (3√3−4)
= =
(3√3+4)(3√3−4) (3√3)2 −(4)2

27+16−24√3 43−24√3
= =
27−16 11
2 ∘ 2 ∘ 2 ∘

(v)
5 cos 60 +4 sec 30 −tan 45

2 ∘ 2 ∘
sin 30 +cos 30

2 2
1 2 2
5( ) + 4( ) − (1)
2 √3
=
2 2
1 √3
( ) + ( )
2 2

1 16
5( ) + ( ) − 1
4 3

=
1 3
+
4 4

15 + 64 − 12 67
= =
4 12
4

Page : 187 , Block Name : Exercise 8.2

Q2 Choose the correct option and justify your choice:
∘
2 tan 30
(i) ∘
=
2
1+tan 30
∘ ∘ ∘ ∘
(A) sin 60 (B) cos 60 (C) tan 60 (D) sin 30
2 ∘
1−tan 45
(ii) ∘
=
2
1+tan 45

∘ ∘
(A) tan 90 (B) 1 (C) sin 45 (D) 0

(iii) sin 2A = 2 sin A is true when A =
∘ ∘ ∘ ∘
(A) 0 (B) 30 (C) 45 (D) 60
∘
2 tan 30
(iv) 2 ∘
=
1−tan 30
∘ ∘ ∘ ∘
(A) cos 60 (B) sin 60 (C) tan 60 (D) sin 30

∘

Answer. (i) 2 tan 30
2 ∘
1+tan 30

1 2
2( )
√3 √3 √3
= = =
2 1 4
1 1+
1+( ) 3 3
√3

6 √3
= =
2
4√3

Out of the given alternatives, only sin 60 ∘ √3
=
2

Hence, (A) is correct.
2 ∘

(ii)
1−tan 45

2 ∘
1+tan 45
2
1−(1) 1−1 0
= 2
= = = 0
1+(1) 1+1 2

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

As sin 2A = sin = 0 ∘

2 sin A = 2 sin = 2(0) = 0 ∘

Hence, (A) is correct.
∘

(iv)
2 tan 30

2 ∘
1−tan 30

1 2 2
2( )
√3 √3 √3
= = =
2 1 2
1 1 −
1 − ( ) 3 3
√3

= √3

Out of the given alternatives, only tan60 ∘
= √3

Hence, (C) is correct.

Page : 187 , Block Name : Exercise 8.2

1
If tan(A + B) = √3 and (A − B) =
Q3 √3

∘ ∘
0 < A + B ≤ 90 , A > B find A and B .

tan(A + B) = √3

⇒ tan(A + B) = tan 60

⇒ A + B = 60 … (1)

1
tan(A − B) =
√3
Answer. 1
⇒ tan(A − B) =
√3

⇒ A − B = 30 … (2)

⇒ 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

Page : 187 , Block Name : Exercise 8.2

State whether the following are true or false. Justify your answer.
Q4
(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) sin θ = cos θ for all values of θ

∘
(v) cot A is not defined for A = 0

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

∘ ∘
Let A = 30 and B = 60

Answer. sin(A + B) = sin(30 ∘
+ 60 )
∘

∘
= sin 90

= 1
∘ ∘
sin A + sin B = sin 30 + sin 60

1 √3 1+√3
= + =
2 2 2

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

∘ 1
sin 30 = = 0.5
2

∘ 1
sin 45 = = 0.707
√2

∘ √3
sin 60 = = 0.866
2

∘ √3
sin 90 = = 0.866
2

Hence, the given statement is true.

(iii) cos 0
∘
= 1

∘ √3
cos 30 = = 0.866
2

∘ 1
cos 45 = = 0.707
√2

∘ 1
cos 60 = = 0.5
2

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

Hence, the given statement is false.

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

∘
This is true when θ = 45

∘ 1
sin 45 =
√2

∘ 1
cos 45 =
√2

It is not true for all other values of θ .

Hence, the given statement is false.
∘
(v) cot A is not defined for A = 0

cos A
cot A =
sin A

∘
∘ cos 0 1
cot 0 = ∘ = = undefined
sin 0 0

Hence, the given statement is true.

Page : 187 , Block Name : Exercise 8.2

Exercise 8.3

Evaluate:
Q1 sin 18
∘
tan 26
∘
∘ ∘ ∘ ∘
(i) ∘
(ii) ∘
(iii) cos 48 − sin 42 (iv) cosec31 − sec 59
cos 72 cot 64

∘ ∘
∘
sin 18 sin(90 −72 )
=
Answer. (i)
∘ ∘
cos 72 cos 72
∘
cos 72
= ∘
= 1
cos 72
∘ ∘
∘
tan 26 tan(90 −64 )
(ii) ∘ = ∘
cot 64 cot 64

∘
cot 64
= ∘
= 1
cot 64

∘ ∘ ∘ ∘ ∘
(iii) cos 48 − sin 42 = cos(90 − 42 ) − sin 42

∘ ∘
= sin 42 − sin 42

= 0

∘ ∘ ∘ ∘ ∘
(iv) cos ec31 − sec 59 = cosec (90 − 59 ) − sec 59

∘ ∘
= sec 59 − sec 59

= 0

Page : 189 , Block Name : Exercise 8.3

Q2 Show that:
∘ ∘ ∘ ∘
(i) tan 48 tan 23 tan 42 tan 67 = 1

∘ ∘ ∘ ∘
(ii) cos 38 cos 52 − sin 38 sin 52 = 0

∘ ∘ ∘ ∘
(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
Answer. ∘ ∘ ∘ ∘
= (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

Page : 189 , Block Name : Exercise 8.3

Q3 If tan 2A = cot(A − 18 ), where 2A is an acute angle, find the value of A.
∘

Given that,
∘
tan 2A = cot(A − 18 )

∘ ∘
cot(90 − 2A) = cot(A − 18 )
Answer. ∘ ∘
90 − 2A = A − 18
∘
108 = 3A
∘
A = 36

Page : 189 , Block Name : Exercise 8.3

Q4 If tan A = cot B, prove that A + B = 90 ∘

Given that,

tan A = cot B

Answer. tan A = tan(90 ∘
− B)

∘
A = 90 − B
∘
A + B = 90

Page : 189 , Block Name : Exercise 8.3

Q5 If sec 4A = cosec (A − 20 ) , where 4A is an acute angle, find the value of A.
∘

Given that,
∘
sec 4A = cosec (A − 20 )

∘ ∘
cosec (90 − 4A) = cosec (A − 20 )
Answer. ∘ ∘
90 − 4A = A − 20
∘
110 = 5A
∘
A = 22

Page : 189 , Block Name : Exercise 8.3

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

Q6 B+C A
sin( ) = cos
2 2

Answer.
We know that for a triangle ABC

∘
∠A + ∠B + ∠C = 180
∘
∠B + ∠C = 180 − ∠A

∠B+∠C ∘ ∠A
= 90 −
2 2

B+C ∘ A
sin( ) = sin(90 − )
2 2

A
= cos( )
2

Page : 190 , Block Name : Exercise 8.3

Q7
∘ ∘ ∘
Express sin 67 + cos 75 in terms of trigonometric ratios of angles between 0 and
∘
45 .

∘ ∘
sin 67 + cos 75

Answer. = sin(90 ∘ ∘
− 23 ) + cos(90
∘ ∘
− 15 )

∘ ∘
= cos 23 + sin 15

Page : 190 , Block Name : Exercise 8.3

Exercise 8.4

Q1 Express the trigonometric ratios sin A sec A and tan A in terms of cot A. ′

Answer. We know that,
2 2
cosec A = 1 + cot A

1 1
=
2 2
cosec A 1+cot A

2 1
sin A =
2
1+cot A

1
sin A = ±
√1+cot2 A

√1 + cot2 A will always be positive as we are adding two positive quantities.
Therefore, sin A = 1

√1+cot2 A

We know that, tan A =
sin A

cos A

However, cot A = cos A

sin A

Therefore, tan A =
1

cot A

Also, sec A = 1 + tan A
2 2

1
= 1 +
2
cot A

2
cot A+1
= 2
cot A

√cot2 A+1

sec A =
cot A

Page : 193 , Block Name : Exercise 8.4

Q2 Write all the other trigonometric ratios of∠A in terms of sec A.

1
cos A =
sec A
2 2
A|so,sin A + cos A = 1

2 2
sin A = 1 − cos A

1
2
sin A = √1 − ( )
2
sec A
Answer. We know that,
2 √sec2 A − 1
sec A − 1
= √ =
2
sec A sec A

2 √sec2 A − 1
sec A
= √ =
2
sec A sec A

2 2
tan A = sec A − 1

2
tan A = √sec A − 1

cos A sec A
cot A = =
sin A sec A

1
=
√sec2 A−1

1 sec A
cosecA = =
sin A √sec2 A−1

Page : 193 , Block Name : Exercise 8.4

Q3 Evaluate
2 ∘ 2 ∘
sin 63 +sin 27
(i) 2 ∘ 2 ∘
cos 17 +cos 73

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

2 ∘ 2 ∘
sin 63 +sin 27

2 ∘ 2 ∘
cos 17 +cos 73

∘ ∘ 2 2 ∘
[sin(90 −27 )] +sin 27

Answer. = [cos(90 −73 )] +cos
∘ ∘ 2
2
73
∘

∘ 2 2 ∘
[cos 27 ] +sin 27
=
∘ 2 2 ∘
[sin 73 ] +cos 73

2 ∘ 2 ∘
cos 27 +sin 27
= 2 ∘ ∘
2
sin 73 +cos 73

1 2 2
= (As sin A + cos A = 1)
1

= 1
∘ ∘ ∘ ∘
(ii) sin 25 cos 65 + cos 25 sin 65

∘ ∘ ∘ ∘ ∘ ∘
= (sin 25 ) {cos(90 − 25 )} + cos 25 {sin(90 − 25 )}

∘ ∘ ∘ ∘
= (sin 25 ) (sin 25 ) + (cos 25 ) (cos 25 )

2 ∘ 2 ∘
= sin 25 + cos 25

2 2
= 1 (As sin A + cos A = 1)

Page : 193 , Block Name : Exercise 8.4

Q4 Choose the correct option. Justify your choice.
(i) 9 sec A − 9 tan A =
2 2

(A) 1 ( B) 9 (C) 8 (D) 0

(ii)(1 + tan θ + sec θ)(1 + cot θ − cosecθ) =

(A) 0 (B) 1 (C) 2 (D) − 1

( iii) (sec A + tan A)(1 − sin A) =

(A) sec A (B) sin A (C) cosecA (D) cos A
2
1+tan A
(iv) =
2
1+cot A

2
(A) sec A (B) − 1
2 2
(C) cot A (D) tan A

2 2
(i) 9sec A − 9 tan A

2 2
= 9 (sec A − tan A)
Answer.
2 2
= 9(1) [As sec A − tan A = 1]

= 9

Hence, alternatives (B) is correct.
(ii) (1 + tan θ + sec θ)(1 + cot θ − cosecθ)
sin θ 1 cos θ 1
= (1 + + ) (1 + − )
cos θ cos θ sin θ sin θ

cos θ + sin θ + 1 sin θ + cos θ − 1
= ( )( )
cos θ sin θ

2 2
(sin θ + cos θ) − (1)
=
sin θ cos θ
2 2
sin θ + cos θ + 2 sin θ cos θ − 1
=
sin θ cos θ

2 sin θ + cos θ + 2 sin θ cos θ − 1
= = 2
sin θ cos θ

2 sin θ cos θ
= = 2
sin θ cos θ

Hence, alternative (C) is incorrect.
(iii) (secA + tanA)(1-sinA)
1 sin A
= ( + ) (1 − sin A)
cos A cos A

1 + sin A
= ( ) (1 − sin A)
cos A

2 2
1 − sin A cos A
= =
cos A cos A

= cos A

Hence, alternative (D) is correct.

2
sin A
2
1+

(iv)
1+tan A cos2 A
=
2 2
1+cot A cos A
1+
2
sin A
2 2
cos A+sin A 1

cos2 A cos2 A
= =
2 1
sin A+cos2 A
2
2 sin A
sin A
2
sin A 2
= 2
= tan A
cos A

Hence, alternative (D) is correct.

Page : 193 , Block Name : Exercise 8.4

Q5
Prove the following identities, where the angles involved are acute angles for which

the expressions are defined.

2 1−cos θ cos A 1+sin A
(i)(cosecθ − cot θ) = (ii) + = 2 sec A
1+cos θ 1+sin A cos A

tan θ cot θ
(iii) + = 1 + sec θcosecθ
1−cot θ 1−tan θ

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

2
1+sec A sin A
(iv) = [ Hint : Simplify LHS and RHS separately]
sec A 1−cos A

cos A−sin A+1 2 2
(v) = cosecA + cot A, using the identity cosec A = 1 + cot A
cos A+sin A−1

1+sin A
(vi)√ = sec A + tan A
1−sin A

3
sin θ−2 sin θ
(vii) 3
= tan θ
2 cos θ−cos θ

2 2 2 2
(viii)(sin A + cosecA) + (cos A + sec A) = 7 + tan A + cot A
1
(ix) (cosecA − sin A)(sec A − cos A) =
tan A+cot A

[Hint : Simplify LHS and RHS separately]

2
2
1+tan A 1−tan A 2
(x) ( 2
) = ( ) = tan A
1+cot A 1−cot A

Answer. (i)(cosecθ − cot θ) 2 1−cos θ
=
1+cos θ

2
L.H.S. = (cosecθ − cot θ)

2
1 cos θ
= ( − )
sin θ sin θ

2 2
(1 − cos θ) (1 − cos θ)
= =
2 2
(sin θ) sin θ
2 2
(1 − cos θ) (1 − cos θ) 1 − cos θ
= = =
2
1 − cos θ (1 − cos θ)(1 + cos θ) 1 + cos θ

= R. H . S.
cos A 1+sin A
(ii) + = 2 sec A
1+sin A cos A

cos A 1 + sin A
L.H.S. = +
1 + sin A cos A
2 2
cos A + (1 + sin A)
=
2
(1 + sin A)(cos A)

2 2
cos A + 1 + sin A + 2 sin A
=
(1 + sin A)(cos A)

2 2
sin A + cos A + 1 + 2 sin A
=
(1 + sin A)(cos A)
1+1+2 sin A 2+2 sin A
= =
(1+sin A)(cos A) (1+sin A)(cos A)

2(1 + sin A) 2
= = = 2 sec A
(1 + sin A)(cos A) cos A

= R. H . S.

(iii)
tan θ cot θ
+ = 1 + sec θcosecθ
1−cot θ 1−tan θ

L.H.S.=
tan θ cot θ
+
1−cot θ 1−tan θ
sin θ cos θ

cos θ sin θ
= +
cos θ sin θ
1− 1−
sin θ cos θ
sin θ

cos θ
=
sin θ−cos θ cos θ
+
sin θ cos θ

2 2
sin θ cos θ
= +
cos θ(sin θ − cos θ) sin θ(sin θ − cos θ)

2 2
1 sin θ cos θ
= [ − ]
(sin θ − cos θ) cos θ sin θ

3 3
1 sin θ − cos θ
= ( )[ ]
sin θ − cos θ sin θ cos θ
2 2
1 (sin θ − cos θ) (sin θ + cos θ + sin θ cos θ)
= ( )[ ]
sin θ − cos θ sin θ cos θ

(1 + sin θ cos θ)
=
(sin θ cos θ)

= secθcosecθ

= R. H . S

2

(iv)
1+sec A sin A
=
sec A 1−cos A
1
1+
1+sec A cos A
L. H. S. = =
1
sec A
cos A

cos A + 1
= = (cos A + 1)
cos A

(1 − cos A)(1 + cos A)
=
(1 − cos A)

2 2
1 − cos A sin A
= =
1 − cos A 1 − cos A

= R.H.S.
(v)
cos A−sin A+1
= cosecA + cot A
cos A+sin A−1

Using the identity cosec A = 1 + cot A 2 2

L.H.S. =
cos A−sin A+1

cos A+sin A−1
cos A sin A 1
− +
sin A sin A sin A
=
cos A sin A 1
+ +
sin A sin A sin A

cot A−1+cosecA
=
cot A+1−cosecA

{(cot A)−(1−cosecA)}{(cot A)−(1−cosecA)}
=
{(cot A)+(1−cosecA)}{(cot A)−(1−cosecA)}

2
(cot A − 1 + cosecA)
=
2 2
(cot A) − (1 − cosecA)

2 2
cot A + 1 + cosec A − 2 cot A − 2cosecA + 2 cot AcosecA
=
2 2
cot A − (1 + cosec A − 2cosecA)
2
2 cosec A + 2 cot A cos ϵcA − 2 cot A − 2 cosec A
=
2 2
cot A − 1 − cosec A + 2cosecA

2cosecA(cosecA + cot A) − 2(cot A + cosecA)
=
2 2
cot A − cosec A − 1 + 2cosecA

(cosecA + cot A)(2cosecA − 2)
=
−1 − 1 + 2cosecA

(cosecA + cot A)(2cosecA − 2)
=
(2cosecA − 2)

= cosecA + cot A

= R. H . S

(vi) √
1+sin A
= sec A + tan A
1−sin A

1 + sin A
L.H.S. = √
1 − sin A

(1 + sin A)(1 + sin A)
= √
(1 − sin A)(1 + sin A)

(1+sin A) 1+sin A
= =
√1−sin 2 √cos2 A
A

1+sin A
= = sec A + tan A
cos A

= R.H.S.
3

(vii)
sin θ−2 sin θ
= tan θ
2 cos θ cos θ
3
sin θ − 2 sin θ
L.H.S. =
3
2 cos θ − cos θ
2
sin θ (1 − 2 sin θ)
=
2
cos θ (2 cos θ − 1)

2
sin θ × (1 − 2 sin θ)
=
2
cos θ × {2 (1 − sin θ) − 1}

2
sin θ × (1 − 2 sin θ)
=
2
cos θ × (1 − 2 sin θ)

= tan θ = R. H . S
2 2 2 2
(viii)(sin A + cosecA) + (cos A + sec A) = 7 + tan A + cot A

L.H.S= (sin A + cosecA) 2
+ (cos A + sec A)
2

2 2 2 2
= sin A + cosec A + 2 sin AcosecA + cos A + sec A + 2 cos A sec A

2 2 2 2 1 1
= (sin A + cos A) + (cosec A + sec A) + 2 sin A ( ) + 2 cos A ( )
sin A cos A

2 2
= (1) + (1 + cot A + 1 + tan A) + (2) + (2)

2 2
= 7 + tan A + cot A

= R. H . S
1
(ix)(cosecA − sin A)(sec A − cos A) =
tan A+cot A

L.H.S = (cosecA − sin A)(sec A − cos A)

1 1
= ( − sin A) ( − cos A)
sin A cos A
2 2
1 − sin A 1 − cos A
= ( )( )
sin A cos A

2 2
(cos A) (sin A)
=
sin A cos A

= sin A cos A

R.H.S=
1

tan A+cot A

1 1
= =
2
sin A cos A sin A+cos2 A
+
cos A sin A sin A cos A

sin A cos A
= = sin A cos A
2 2
sin A + cos A

Hence, L.H.S = R.H.S
2
2
1+tan A 1−tan A 2
(x) ( ) = ( ) = tan A
2
1+cot A 1−cot A

2 2 2
sin A cos A+sin A
2 1+
1+tan A cos2 A cos2 A

2
= =
1+cot A cos2 A cos2 A
1+
2 2
sin A sin A

1
2
cos
2
A sin A
= =
1 2
cos A
2
sin A

2
=tan A

2 2
1 − tan A 1 + tan A − 2 tan A
( ) =
2
1 − cot A 1 + cot A − 2 cot A
2
sec A − 2 tan A
=
2
cos ec A − 2 cot A
1 2 sin A 1−2 sin A cos A
−
cos2 A cos A cos2 A
= =
1 2 cos A 2
1−2 sin A
−
sin2 A sin A
sin2 A
2
sin A 2
= 2
= tan A
cos A

Page : 193 , Block Name : Exercise 8.4