# 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 error: Content is protected !! 