R S AGGARWAL AND V AGGARWAL Solutions Mathematics Class 10 Chapter 7 Trigonometric Ratios of Complementary Angles

Chapter 7 – Trigonometric Ratios of Complementary Angles Exercise Ex. 7

Question 1

Without using trigonometric tables, evaluate:

Solution 1

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Question 2

Without using trigonometric tables, prove that:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Solution 2

(i)                LHS = cos81° – sin9°

= cos(90° -9°)- sin9° = sin9° – sin9°

= 0 = RHS

(ii)              LHS = tan71° – cot19°

=tan(90° – 19°) – cot19° =cot19° – cot19°

=0 = RHS

(iii)               LHS = cosec80° – sec10°

= cosec(90° – 10°) – sec(10°)

= sec10° – sec10° = 0

= RHS

(iv)               LHS=

(v)

(vi)

(vii)

LHS = RHS

(viii)

(ix)              LHS = (sin65° + cos25°) (sin65° – cos25°)

Question 3

Without
using trigonometric tables, prove that:

sin
53
cos37
+ cos53 sin37
= 1

Solution 3

Question 4

cos
54
cos 36– sin 54
sin36
= 0

Solution 4

Question 5

sec
70
sin 20
+ cos 20
cosec 70 = 2

Solution 5

Question 6

sin
35
sin 55
– cos 35
cos 55
= 0

Solution 6

Question 7

(sin
72
+ cos 18)(sin72
– cos18)
= 0

Solution 7

Question 8

tan
48
tan 23
tan 42
tan 67
= 1

Solution 8

Question 9

Prove
that:

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Prove
that:

sin
θ
cos (90
– θ)
+ sin (90
– θ) cos θ = 1

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

cos
1
cos2
cos3
… cos 180=
0

Solution 24

Question 25

Solution 25

Question 26

Prove that:

(i)

(ii)

(iii)

(iv)

(v)

Solution 26

(i)                  LHS =

= RHS

(ii)                LHS =

= RHS

(iii)                  LHS =

= RHS

(iv)             LHS = cosec(65° + ) – sec(25° ) – tan(55° ) + cot(35° +  )

= RHS

(v)        LHS =

= 0 + 1 = 1

= RHS

Question 27

Express
each of the following in terms of T-ratios of angles lying between 0
and 45:

sin
67
+ cos 75

Solution 27

Question 28

cot
65
+ tan 49

Solution 28

Question 29

sec
78
+ cosec 56

Solution 29

Question 30

cosec
54
+ sin 72

Solution 30

Question 31

If A, B, C are the angles of a triangle ABC, prove that:

Solution 31

A + B + C = 180°

So, B + C= 180° – A

Question 32

If
cos 2θ
= sin 4θ,
where 2θ
and 4θ
are acute angles, find the value of θ.

Solution 32

Question 33

If
sec 2A= cosec (A – 42ᵒ), where 2A is an
acute angle, find the value of A.

Solution 33

Question 34

If sin3A = cos(A – 26o), where 3A is an acute angle, find the value of A.

Solution 34

Question 35

If tan 2A = cot(A – 12o), where 2A is an acute angle, find the value of A.

Solution 35

Question 36

If sec 4A = cosec(A – 15o), where 4A is an acute angle, find the value of A.

Solution 36

Question 37

Prove
that:

Solution 37

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