 Lesson: Introduction to Trigonometry
Exercise 8.1 (11)
Question: 1
In ABC right-angled at B, AB 24 cm, BC 7 cm.
Determine:
(i)
sinA,
cosA
(ii)
sinC,
cosC
Solution
By the Pythagoras theorem in triangle ABC,
2 2 2
AC AB BC
2 2 2
AC 24 7
2
AC 576 49
AC 625
Fig. Exc_8.1_1 (i)
sideoppositeto A BC
sin A
hypotenuse AC
(ii)
7
sin A
25
cosA
Hypotenuse AC
24
cosA
25
(iii)
sideoppositeto C AB
sin C
hypotenuse AC
24
sin C
25
cosC
Hypotenuse AC
7
cosC
25
Question: 2
In Fig. 8.13, find . Fig. Exc_8.1_2
Solution
By the Pythagoras theorem in ,
PQR
2 2 2
PR PQ QR
 
2 2 2
13 12 QR
2
QR 169 144
2
QR 25
QR 5
sideoppositeto P QR
tanP
5
tanP
12
cotR
sideoppositeto R PQ 5
12
5 5
tanP cotR
12 12
tanP cotR 0
Question: 3
If , calculate and .
3
sin A
4
cosA
tanA
Solution
Let us consider a triangle ABC, right-angled at B.
Fig. Exc_8.1_3 BC 3
sin A
AC 4
Let .
BC 3a
Thus, .
AC 4a
By the Pythagoras theorem in triangle ABC,
2 2 2
AC AB BC
2 2 2
4a AB 3a
2
2 2
16a 9a AB
2
2
AB 7a
AB
cosA
AC
7a
4a
7
4
BC
tanA
AB
3a
7a
3
7
Hence, and .
7
cosA
4
3
tanA
7 Question: 4
Given , find and .
15cot A 8
sinA
secA
Solution
Let us consider a triangle ABC, right-angled at B.
.
AB 8
cot A
BC 15
Let and .
AB 8a
BC 15a
By the Pythagoras theorem
2 2 2
AC AB BC
2 2 2
AC 8a 15a
2
2 2
AC 64a 225a
2
2
AC 289a
AC 17a
Fig. Exc_8.1_4 BC
sinA
AC
15a
17a
15
17
AC
secA
AB
17
8
Hence, and .
15
sinA
17
17
secA
8
Question: 5
Given , calculate all other trigonometric
13
secθ
12
ratios.
Solution
Let us consider a triangle ABC, right-angled at B.
AC 13
secθ
AB 12 Fig. Exc_8.1_5
Let and .
AC 13a
AB 12a
By the Pythagoras theorem,
2 2 2
AC AB BC
2 2 2
13a 12a BC
2
2 2
BC 169a 144a
2
2
BC 25a
2
BC 25a
BC 5a
BC 5a 5
sinθ
AC 13a 13
AB 12a 12
cosθ
AC 13a 13
BC 5a 5
tanθ
AB 12a 12 AB 12a 12
cot θ
BC 5a 5
AC 13a 13
cosecθ
BC 5a 5
Question: 6
If and are acute angles such that
A
B
, then show that .
cosA cosB
A B
Solution
Let us consider a triangle ABC such that CD AB.
cosA cosB
AC BC
BD BC
Let .
a
BD BC
AC aBC ......(2) Fig. Exc_8.1_6
In triangles CAD and CBD, by the Pythagoras
theorem,
2 2 2
2 2 2
CD BC BD ......(4)
From equations (3) and (4).
2 2 2 2
2 2 2 2
aBC aBD BC BD
2 2 2 2
2
a BC BD BC BD
2
a 1
a 1
Put the value of in equation (2).
a
AC BC Since the angles opposite to equal sides in a triangle
are equal.
Thus,
.
A B
Question: 7
If , evaluate:
7
cot θ
8
(i)
1 sinθ 1 sinθ
1 cosθ 1 cosθ
(ii)
2
cot θ
Solution
Let us consider a triangle ABC, right-angled at B.
BC 7
cot θ
AB 8
Fig. Exc_8.1_7
Let and .
BC 7a
AB 8a In triangle ABC, by the Pythagoras theorem,
2 2 2
AC AB BC
2 2 2
AC 8a 7a
2
2 2
AC 64a 49a
2
2
AC 113a
AC 113a
AB
sinθ
AC
8a
113a
8
113
BC
cosθ
AC
7a
113a
7
113
(i)
1 sinθ 1 sinθ
1 cosθ 1 cosθ
2
2
1 sin θ
1 sinθ 1 sinθ
1 cosθ 1 cosθ
1 cos θ 2
2
8
1
113
7
1
113
64
1
113
49
1
113
49
113
64
113
49
64
Thus,
1 sinθ 1 sinθ
49
1 cosθ 1 cosθ 64
(ii)
2
cot θ
2
2
cot θ cot θ
2
7
8
49
64
Thus, .
2
49
cot θ
64 Question: 8
If , check whether
3cot A 4
or not.
2
2 2
2
1 tan A
cos A sin A
1 tan A
Solution
Given:
3cot A 4
4
cot A
3
Let us consider a triangle ABC, right-angled at B.
Fig. Exc_8.1_8
AB 4
cot A
BC 3
Let and .
AB 4a
BC 3a
By the Pythagoras theorem,
2 2 2
AC AB BC
2 2 2
AC 4a 3a 2
2 2
AC 16a 9a
2
2
AC 25a
2
AC 25a
AC 5a
AB
cosA
AC
4a
5a
4
5
BC
sinA
AC
3a
5a
3
5
BC
tanA
AB
3a
4a
3
4
Now, find the value of .
2
2
1 tan A
1 tan A 2
2
2
2
3
1
1 tan A
4
1 tan A
3
1
4
9
1
16
9
1
16
16 9
16
16 9
16
7
25
Now, find the value of .
2 2
cos A sin A
2 2
2 2
4 3
cos A sin A
5 5
16 9
25 25
7
25
Hence, .
2
2 2
2
1 tan A
cos A sin A
1 tan A
Question: 9 In triangle ABC, right-angled at B, if , find
1
tanA
3
the value of:
(i)
sinAcosC cosAsinC
(ii)
cosAcosC sinAsinC
Solution
In the figure, ABC is a triangle, right-angled at B.
Fig. Exc_8.1_9
BC 1
tanA
AB
3
Let and .
BC a
AB 3a
By the Pythagoras theorem,
2 2 2
AC AB BC
2
2
2
AC 3a a
2
2 2
AC 3a a 2
2
AC 4a
2
AC 4a
AC 2a
BC
sinA
AC
a
2a
1
2
AB
cosA
AC
3a
2a
3
2
AB
sinC
AC
3a
2a
3
2
BC
cosC
AC
a
2a 1
2
(i) Given:
sinAcosC cosAsinC
1 1 3 3
sinAcosC cosAsinC
2 2 2 2
1 3
4 4
1
Hence, .
sinAcosC cosAsinC 1
(ii) Given:
cosAcosC sinAsinC
3 1 1 3
cosAcosC sinAsinC
2 2 2 2
3 3
4 4
0
Hence, .
cosAcosC sinAsinC 0
Question: 10
In PQR, right-angled at Q, PR QR 25 cm and
PQ 5 cm. Determine the values of and
sinP,cosP
.
tanP
Solution
Given: PR QR 25 cm and PQ 5 cm.
Let .
PR x
Thus, .
QR 25 x
Fig. Exc_8.1_10
In , by the Pythagoras theorem,
PQR
2 2 2
PR PQ QR
2 2
2
x 5 25 x
2 2
x 25 625 x 2 25 x
50x 25 625
650
x
50
x 13 Thus, .
PR 13cm
QR 25 13 12cm
QR 12
sinP
PR 13
PQ 5
cosP
PR 13
QR 12
tanP
PQ 5
Question: 11
State whether the following are true or false. Justify
(i) The value of is always less than 1.
tanA
(ii) for some value of angle A.
12
secA
5
(iii) is the abbreviation used for the cosecant
cosA
of angle A.
(iv) is the product of cot and A.
cot A
(v) for some angle .
4
sinθ
3
θ
Solution
(i) False.
Let us consider a triangle ABC, which is right-
angled at B. Let and .
BC 12
AB 5
Fig. Exc_8.1_11_(i)
BC 12
tanA
AB 5
12
tanA 1
5
Hence, the value of is not always less
tanA
than 1.
(ii) True.
Let us consider a triangle ABC, which is right-
angled at B. Fig. Exc_8.1_11_(ii)
AC 12
secA
AB 5
Let and .
AC 12a
AB 5a
By the Pythagoras theorem.
2 2 2
AC AB BC
2 2 2
12a 5a BC
2
2 2
BC 144a 25a
2
2
BC 119a
BC 10.9a ......(1)
But, BC should satisfy the condition,
AC AB BC AC AB
12a 5a BC 12a 5a
12a 5a BC 12a 5a12a 5a BC 12a 5a
From equation (1). BC 10.9a
Hence, this type of triangle can be drawn and
the given statement is true.
(iii) False.
is the abbreviation used for cosecant of
cosec A
angle A and the abbreviation used for cosine of
angle A is .
cosA
(iv) False.
cot A is the cotangent of .
A
(v) False.
sideoppositeto θ
sinθ
hypotenuse
We know that in a right-angled triangle,
hypotenuse is greater than the other two sides.
Thus, such value of is impossible.
sinθ
Exercise 8.2 (4)
Question: 1
Evaluate the following: (i)
sin60 cos30 sin30 cos60
(ii)
2 2 2
2tan 45 cos 30 sin 60
(iii)
cos45
sec30 cosec30
(iv)
sin30 tan45 cosec60
sec30 cos60 cot 45
(v)
2 2 2
2 2
5cos 60 4sec 30 tan 45
sin 30 cos 30
Solution
(i) Given:
sin60 cos30 sin30 cos60
3 3 1 1
sin60 cos30 sin30 cos60
2 2 2 2
3 1
4 4
4
4
1
Thus,
sin60 cos30 sin30 cos60 1
(ii)
2 2 2
2tan 45 cos 30 sin 60
2 2
2
2 2 2
3 3
2tan 45 cos 30 sin 60 2 1
2 2
 
3 3
2
4 4
2
Thus,
2 2 2
2tan 45 cos 30 sin 60 2 (iii) Given:
cos45
sec30 cosec30
1
cos45
2
2
sec30 cosec30
2
3
1
2
2 2 3
3
3
2 2 2 3
3
2 2 2 6
Multiply numerator and denominator by
.
2 6 2 2
cos45 3 2 6 2 2
sec30 cosec30
2 2 2 6 2 6 2 2
2 2
2 3 6 2
2 6 2 2
2 3 6 2
24 8 2 3 6 2
16
18 6
8
3 2 6
8
Thus, .
cos45 3 2 6
sec30 cosec30 8
(iv) Given:
sin30 tan45 cosec60
sec30 cos60 cot 45
1 2
1
sin30 tan45 cosec60
2
3
2 1
sec30 cos60 cot 45
1
2
3
3 2
2
3
3 2
2
3
3 3 4
2 3
3 3 4
2 3
3 3 4
3 3 4 Multiply both numerator and denominator by
.
3 3 4
sin30 tan45 cosec60 3 3 4 3 3 4
sec30 cos60 cot 45
3 3 4 3 3 4
2
2
2
3 3 4
3 3 4
27 16 24 3
27 16
43 24 3
11
Thus,
sin30 tan45 cosec60 43 24 3
sec30 cos60 cot 45 11
(v) Given:
2 2 2
2 2
5cos 60 4sec 30 tan 45
sin 30 cos 30
2
2
2
2 2 2
2
2 2
2
1 2
5 4 1
5cos 60 4sec 30 tan 45
2
3
sin 30 cos 30
1 3
2 2
1 16
5 1
4 3
1 3
4 4 15 64 12
12
4
4
67
12
Thus,
.
2 2 2
2 2
5cos 60 4sec 30 tan 45 67
sin 30 cos 30 12
Question: 2
Choose the correct option and justify your choice:
(i)
2
2tan30
1 tan 30
(A)
sin60
(B)
cos60
(C)
tan60
(D)
sin30
(ii)
2
2
1 tan 45
1 tan 45
(A)
tan90
(B) 1
(C)
sin45
(D) 0
(iii) is true when
sin2A 2sinA
A
(A)
0
(B)
30
(C)
45
(D)
60 (iv)
2
2tan30
1 tan 30
(A)
cos60
(B)
sin60
(C)
tan60
(D)
sin30
Solution
(i)
2
2tan30
1 tan 30
(A)
2
2
1
2
2tan30
3
1 tan 30
1
1
3
2
3
1
1
3
2
3
4
3
3 2
4 3
3
2
We know that .
3
sin60
2 (ii)
2
2
1 tan 45
1 tan 45
(D)
2
2
2
2
1 1
1 tan 45
1 tan 45
1 1
1 1
1 1
0
(iii)
sin2A 2sinA
(A)
Put .
A 0
sin2A sin2 0
sin0
0
2sinA 2sin0
2 0
0
is true when .
sin2A 2sinA
A 0
(iv)
2
2tan30
1 tan 30
(C) 2
2
1
2
2tan30
3
1 tan 30
1
1
3
2
3
1
1
3
2
3
3 1
3
2
3
2
3
3
We know that .
tan60 3
Thus, .
2
2tan30
tan60
1 tan 30
Question: 3
If
tan A B 3
1
tan A B
3
; , find A and B.
0 A B 90
A B Solution
tan A B 3
We know that .
tan60 3
tan A B tan60
Thus,
A B 60 ......(1)
1
tan A B
3
tan A B tan30
Thus,
A B 30 ......(2)
A B A B 60 30
2A 90
90
A
2
A 45
Put the value of A in equation (1).
A B 60
45 B 60
B 60 45
B 15
Thus, and .
A 45
B 15
Question: 4 State whether the following are true or false. Justify
(i)
sin A B sinA sinB
(ii) The value of increases as increases.
sinθ
θ
(iii) The value of increases as increases.
cosθ
θ
(iv) for all values of .
sinθ cosθ
θ
(v) is not defined for
cot A
A 0
Solution
(i) False.
sin A B sinA sinB
Let and .
A 30
B 60
sin A B sin 30 60
sin90
1
sinA sinB sin30 sin60
1 3
2 2
1 3
2
Hence, .
sin A B sinA sinB
(ii) True.
Find the value of for increasing values of .
sinθ
θ
Let and .
θ 30 ,
45
60 1
sin30 0.5
2
1
sin45 0.707
2
3
sin60 0.866
2
sin90 1
Thus, the value of increases as increases.
sinθ
θ
(iii) False.
Find the value of for increasing values of
cosθ
θ
.
Let and .
θ 30 ,
45
60
3
cos30 0.866
2
1
cos45 0.707
2
1
cos60 0.5
2
3
cos30 0.866
2
1
cos45 0.707
2
1
cos60 0.5
2 Hence, the value of does not increase as
cosθ
θ
increases.
(iv) False.
Find and for some values of .
sinθ
cosθ
θ
Let and .
θ 45
30
θ 45 and30
and
1
sin45
2
1
cos45
2
and
1
sin30
2
3
cos30
2
Hence, for all values of .
sinθ cosθ
θ
(v) True.
cosA
cot A
sinA
Let .
A 0
cos0
cot 0
sin0
 
1
undefined
0
Hence, is not defined for
cot A
A 0
Exercise 8.3 (7) Question: 1
Evaluate:
(i)
sin18
cos72
(ii)
tan26
cot 64
(iii)
cos48 sin42
(iv)
cosec31 sec59
Solution
(i) Given:
sin18
cos72
sin 90 72
sin18
cos72 cos72
cos72
sin 90 72 cos72
cos72
1
Hence, .
sin18
1
cos72
(ii) Given:
tan26
cot 64
tan 90 64
tan26
cot 64 cot 64
cot 64
tan 90 64 cot 64
cot 64
Hence, .
1
tan26
1
cot 64 (iii) Given:
cos48 sin42
cos48 sin42 cos 90 42 sin42
sin42 sin42 cos 90 42 sin42
0
Hence, .
cos48 sin42 0
(iv) Given:
cosec31 sec59
cosec31 sec59 cosec 90 59 sec59
sec59 sec59 cosec 90 59 sec59
0
Hence, .
cosec31 sec59 0
Question: 2
Show that:
(i)
tan48 tan23 tan42 tan67 1
(ii)
cos38 cos52 sin38 sin52 0
Solution
(i) Given:
tan48 tan23 tan42 tan67 1
We know that .
tan 90 θ cot θ
tan 90 42
tan48 tan23 tan42 tan67 tan 90 67
tan42 tan67
cot 42 cot 67 tan42 tan67 cot 42 tan42 cot 67 tan67
1 1
cot 42 cot 67
cot 42 cot 67
1
Thus, .
tan48 tan23 tan42 tan67 1
(ii) Given:
cos38 cos52 sin38 sin52 0
We know that .
cos 90 θ sinθ
cos 90 52
cos38 cos52 sin38 sin52 cos 90 38
sin38 sin52
sin52 sin38 sin38 sin52
0
Thus, .
cos38 cos52 sin38 sin52 0
Question: 3
If ,where is an acute angle,
tan2A cot A 18
2A
find the value of A.
Solution
Given:
tan2A cot A 18
tan2A cot A 18
cot 90 2A cot A 18 90 2A A 18
2A A 90 18
3A 108
108
A
3
A 36
Hence, A is .
36
Question: 4
If , prove that .
tanA cotB
A B 90
Solution
Given:
tanA cotB tan 90 B cotB
tanA tan 90 B
A 90 B
A B 90
Hence proved.
Question: 5
If , where 4A is an acute
sec4A cosec A 20
angle, find the value of A.
Solution Given:
sec4A cosec A 20
sec4A cosec A 20
cosec 90 4A cosec A 20 cosec 90 4A sec4A
90 4A A 20
4A A 90 20
5A 110
110
A
5
A 22
Hence, .
A 22
Question: 6
If A, B and C are interior angles of a triangle ABC,
then show that
.
B C A
sin cos
2 2
Solution
By the angle sum property of a triangle,
A B C 180
B C 180 A
Divide both sides by 2.
B C 180 A
2 2 B C A
90
2 2
Take sin of angles on both sides.
B C A
sin sin 90
2 2
A
cos
2
Hence, .
B C A
sin cos
2 2
Question: 7
Express in terms of trigonometric
sin67 cos75
ratios of angles between and .
0
45
Solution
Given:
sin67 cos75
sin67 cos75 sin 90 23 cos 90 15
cos23 sin15
Hence,
sin67 cos75 cos23 sin15
Exercise 8.4 (5)
Question: 1 Express the trigonometric ratios sin A, sec A and tan
A in terms of cot A.
Solution
We know that,
2 2
cosec A 1 cot A
2 2
1 1
cosec A 1 cot A
2
2
1 1
sin A sinA
1 cot A cosec A
Thus,
2
1
sinA
1 cot A
We know that,
and
sinA
tanA
cosA
cosA
cot A
sinA
Thus,
1
tanA
cot A
We know that,
2 2
sec A 1 tan A
2
2
1
sec A 1
cot A
2
2
2
cot A 1
sec A
cot A
2
cot A 1
secA
cot A
Thus, 2
1
sinA
1 cot A
and .
2
cot A 1
secA
cot A
1
tanA
cot A
Question: 2
Write all the other trigonometric ratios of in
A
terms of sec A.
Solution
We know that,
1
cosA
secA
Now,
2 2
sin A cos A 1
2 2
sin A 1 cos A
2
sinA 1 cos A
2
1
sinA 1
secA
2
2
sec A 1
sinA
sec A
2
sec A 1
sinA ......(1)
secA Now, we know that,
2 2
tan A 1 sec A
2 2
tan A sec A 1
Now, for ,
cot A
cosA
cot A
sinA
From equation (1).
2
tanA sec A 1
cosA
cot A
sinA
2
1
secA
sec A 1
secA
2
1
sec A 1
We know that .
1
cosec A
sinA
From equation (1).
1
cosec A
sinA
2
secA
sec A 1
Question: 3
Evaluate: (i)
2 2
2 2
sin 63 sin 27
cos 17 cos 73
(ii)
sin25 cos65 cos25 sin65
Solution
(i) Given:
2 2
2 2
sin 63 sin 27
cos 17 cos 73
2
2
2 2
2
2 2
2
sin 90 27 sin 27
sin 63 sin 27
cos 17 cos 73
cos 90 73 cos 73
2
2
2
2
cos27 sin 27
sin73 cos 73
2 2
2 2
cos 27 sin 27
sin 73 cos 73
1
1
1
Hence, .
2 2
2 2
sin 63 sin 27
1
cos 17 cos 73
(ii) Given:
sin25 cos65 cos25 sin65
sin25 cos 90 25
sin25 cos65 cos25 sin65
cos25 sin 90 25
sin25 sin25 cos25 cos25
2 2
sin 25 cos 25
1
Hence, .
sin25 cos65 cos25 sin65 1
Question: 4 Choose the correct option. Justify your choice.
(i)
2 2
9sec A 9tan A
(A) 1
(B) 9
(C) 8
(D) 0
(ii)
1 tanθ secθ 1 cot θ cosecθ
(A) 0
(B) 1
(C) 2
(D)
1
(iii)
secA tanA 1 sinA
(A)
secA
(B)
sinA
(C)
cosec A
(D)
cosA
(iv)
2
2
1 tan A
1 cot A
(A)
2
sec A
(B)
1
(C)
2
cot A
(D)
2
tan A
Solution
(i) Given:
2 2
9sec A 9tan A
(B)
2 2 2 2
9sec A 9tan A 9 sec A tan A
2 2
9 1 sec A tan A 1 9
Thus, .
2 2
9sec A 9tan A 9
(ii) Given:
1 tanθ secθ 1 cot θ cosecθ
(C)
sinθ 1
1
cosθ cosθ
1 tanθ secθ 1 cot θ cosecθ
cosθ 1
1
sinθ sinθ
cosθ sinθ 1 sinθ cosθ 1
cosθ sinθ
2 2
sinθ cosθ 1
sinθcosθ
2 2
sin θ cos θ 2sinθcosθ 1
sinθcosθ
We know that .
2 2
sin θ cos θ 1
Thus,
1 2sinθcosθ 1
1 tanθ secθ 1 cot θ cosecθ
sinθcosθ
2sinθcosθ
sinθcosθ
2
Hence,
1 tanθ secθ 1 cot θ cosecθ 2
(iii) Given:
secA tanA 1 sinA (D)
1 sinA
secA tanA 1 sinA 1 sinA
cosA cosA
1 sinA
1 sinA
cosA
2
1 sin A
cosA
2
2 2
cos A
1 sin A cos A
cosA
cosA
Hence,
secA tanA 1 sinA cosA
(iv) Given:
2
2
1 tan A
1 cot A
(D)
2
2
2
2
2
2
sin A
1
1 tan A
cos A
cos A
1 cot A
1
sin A
2 2
2
2 2
2
cos A sin A
cos A
sin A cos A
sin A
2
2
sin A
cos A
2
tan A Hence, .
2
2
2
1 tan A
tan A
1 cot A
Question: 5
Prove the following identities, where the angles
involved are acute angles for which the expressions
are defined.
(i)
2
1 cosθ
cosecθ cot θ
1 cosθ
(ii)
cosA 1 sinA
2secA
1 sinA cosA
(iii)
tanθ cot θ
1 secθ cosecθ
1 cot θ 1 tanθ
(iv)
2
1 secA sin A
secA 1 cosA
(v)
cosA sinA 1
cosec A cot A
cosA sinA 1
, using the identity
2 2
cosec A 1 cot A
(vi)
1 sinA
secA tanA
1 sinA
(vii)
3
3
sinθ 2sin θ
tanθ
2cos θ cosθ
(viii)
2 2
2 2
sinA cosec A cosA secA 7 tan A cot A
(ix)
1
cosec A sinA secA cosA
tanA cot A
(x)
2
2
2
2
1 tan A 1 tanA
tan A
1 cot A 1 cot A Solution
(i) Given:
2
1 cosθ
cosecθ cot θ
1 cosθ
LHS
2
2
1 cosθ
cosecθ cot θ
sinθ sinθ
2
2
1 cosθ
sin θ
2
2 2
2
1 cosθ
1 cos θ sin θ
1 cos θ
1 cosθ 1 cosθ
1 cosθ 1 cosθ
1 cosθ
1 cosθ
RHS
Hence proved.
(ii) Given:
cosA 1 sinA
2secA
1 sinA cosA
LHS
cosA cosA 1 sinA 1 sinA
cosA 1 sinA
1 sinA cosA 1 sinA cosA
2 2
cos A 1 sin A 2sinA
1 sinA cosA 2 2
sin A cos A 1 2sinA
1 sinA cosA
2 2
1 1 2sinA
sin A cos A 1
1 sinA cosA
2 1 sinA
1 sinA cosA
2
cosA
2secA
RHS
Hence proved.
(iii) Given:
tanθ cot θ
1 secθ cosecθ
1 cot θ 1 tanθ
LHS
sinθ cosθ
tanθ cot θ
cosθ sinθ
cosθ sinθ
1 cot θ 1 tanθ
1 1
sinθ cosθ
sinθ cosθ
cosθ sinθ
sinθ cosθ cosθ sinθ
sinθ cosθ
2 2
sin θ cos θ
cosθ sinθ cosθ sinθ sinθ cosθ
2 2
1 sin θ cos θ
sinθ cosθ cosθ sinθ 3 3
1 sin θ cos θ
sinθ cosθ sinθcosθ
2 2
sinθ cosθ sin θ cos θ sinθcosθ
1
sinθ cosθ sinθcosθ
2 2
1 sinθcosθ
sin θ cos θ 1
sinθcosθ
1 sinθcosθ
sinθcosθ sinθcosθ
secθ cosecθ 1
RHS
Hence proved.
(iv) Given:
2
1 secA sin A
secA 1 cosA
LHS
1
1
1 secA
cosA
1
secA
cosA
cosA 1
cosA
1
cosA
cosA 1
Multiply both numerator and denominator by
.
1 cosA 2
1 cosA 1 cosA
1 cos A
1 cosA 1 cosA
2
2 2
sin A
1 cos A sin A
1 cosA
RHS
Hence proved.
(v) Given:
cosA sinA 1
cosec A cot A
cosA sinA 1
Divide both numerator and denominator in LHS
by .
sinA
LHS
cosA sinA 1
cosA sinA 1
sinA sinA sinA
cosA sinA 1
cosA sinA 1
sinA sinA sinA
cot A 1 cosec A
cot A 1 cosec A
Multiply both numerator and denominator by
.
cot A 1 cosec A
cot A 1 cosec A cot A 1 cosec A
cot A 1 cosec A cot A 1 cosec A
2
2 2
cot A 1 cosec A
cot A 1 cosec A 2 2
2 2
cot A 1 cosec A 2cot A 2cosec A 2cot Acosec A
cot A 1 cosec A 2cosec A
2 2
2cosec A cosec A cot A 2 cot A cosec A
cot A cosec A 1 2cosec A
cosec A cot A 2cosec A 2
1 1 2cosec A
cosec A cot A 2cosec A 2
2cosec A 2
cosec A cot A
Hence proved.
RHS
(vi) Given:
1 sinA
secA tanA
1 sinA
Multiply both numerator and denominator in
LHS by .
1 sinA
LHS
1 sinA 1 sinA
1 sinA
1 sinA 1 sinA 1 sinA
2
1 sinA
1 sin A
2 2
2
1 sinA
1 sin A cos A
cos A
1 sinA
cosA
1 sinA
cosA cosA secA tanA
RHS
Hence proved.
(vii) Given:
3
3
sinθ 2sin θ
tanθ
2cos θ cosθ
LHS
2
3
3
2
sinθ 1 2sin θ
sinθ 2sin θ
2cos θ cosθ
cosθ 2cos θ 1
2
2 2
2
sinθ 1 2sin θ
1 sin θ cos θ
cosθ 2 1 sin θ 1
2
2
sinθ 1 2sin θ
cosθ 1 2sin θ
tanθ
RHS
Hence proved.
(viii) Given:
2 2
2 2
sinA cosec A cosA secA 7 tan A cot A
LHS 2 2
sinA cosec A cosA secA
2 2
2 2
sin A cosec A 2sinAcosec A
cos A sec A 2cosAsecA
2 2 2 2
sin A cos A cosec A sec A
1 1
2sinA 2cosA
sinA cosA
2 2
1 1 cot A 1 tan A 2 2
2 2
7 tan A cot A
RHS
Hence proved.
(ix) Given:
1
cosec A sinA secA cosA
tanA cot A
Given:
1 1
cosec A sinA secA cosA sinA cosA
sinA cosA
2 2
1 sin A 1 cos A
sinA cosA
2 2
cos Asin A
sinAcosA
sinAcosA
Now take RHS. 1 1
sinA cosA
tanA cot A
cosA sinA
2 2
1
sin A cos A
sinAcosA
2 2
sinAcosA
sin A cos A
2 2
sinAcosA sin A cos A 1
Thus, LHS RHS.
Hence proved.
(x) Given:
2
2
2
2
1 tan A 1 tanA
tan A
1 cot A 1 cot A
LHS
2
2
2
2
2
2
sin A
1
1 tan A
cos A
cos A
1 cot A
1
sin A 2 2
2
2 2
2
cos A+sin A
cos A
sin A cos A
sin A
2
2 2
2
1
cos A
sin A cos A 1
1
sin A
2
2
sin A
cos A
2
tan A
Now take middle term.
2
2
2
1 tanA 1 tan A 2tanA
1 cot A 1 cot A 2cot A
2 2
2
2
2 2
1 tan A sec A
sec A 2tanA
cosec A 2cot A
1 cot A cosec A
2
2
1 2sinA
cos A cosA
1 2cosA
sin A sinA
2
2
1 2sinAcosA
cos A
1 2sinAcosA
sin A
2
2
sin A
cos A
2
tan A Thus, .
2
2
2
2
1 tan A 1 tanA
tan A
1 cot A 1 cot A
Hence proved.