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Introduction to Trigonometry
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NCERT Solutions
Introduction to Trigonometry
19 Solutions
Exercise:
All Exercises
EXERCISE 8.1
EXERCISE 8.2
EXERCISE 8.3
Q1
EXERCISE 8.1
In
△
A
B
C
\triangle \mathrm{ABC}
△
ABC
, right-angled at
B
,
A
B
=
24
c
m
,
B
C
=
7
c
m
\mathrm{B}, \mathrm{AB}=24 \mathrm{~cm}, \mathrm{BC}=7 \mathrm{~cm}
B
,
AB
=
24
cm
,
BC
=
7
cm
. Determine :
(i)
sin
A
,
cos
A
\sin \mathrm{A}, \cos \mathrm{A}
sin
A
,
cos
A
(ii)
sin
C
,
cos
C
\sin \mathrm{C}, \cos \mathrm{C}
sin
C
,
cos
C
Q2
EXERCISE 8.1
In Fig. 8.13, find
tan
P
−
cot
R
\tan \mathrm{P}-\cot \mathrm{R}
tan
P
−
cot
R
.
Q3
EXERCISE 8.1
If
sin
A
=
3
4
\sin \mathrm{A}=\frac{3}{4}
sin
A
=
4
3
, calculate
cos
A
\cos \mathrm{A}
cos
A
and
tan
A
\tan \mathrm{A}
tan
A
.
Q4
EXERCISE 8.1
Given
15
cot
A
=
8
15 \cot \mathrm{A}=8
15
cot
A
=
8
, find
sin
A
\sin \mathrm{A}
sin
A
and
sec
A
\sec \mathrm{A}
sec
A
.
Q5
EXERCISE 8.1
Given
sec
θ
=
13
12
\sec \theta=\frac{13}{12}
sec
θ
=
12
13
, calculate all other trigonometric ratios.
Q6
EXERCISE 8.1
If
∠
A
\angle \mathrm{A}
∠
A
and
∠
B
\angle \mathrm{B}
∠
B
are acute angles such that
cos
A
=
cos
B
\cos \mathrm{A}=\cos \mathrm{B}
cos
A
=
cos
B
, then show that
∠
A
=
∠
B
\angle \mathrm{A}=\angle \mathrm{B}
∠
A
=
∠
B
.
Q7
EXERCISE 8.1
If
cot
θ
=
7
8
\cot \theta=\frac{7}{8}
cot
θ
=
8
7
, evaluate :
(i)
(
1
+
sin
θ
)
(
1
−
sin
θ
)
(
1
+
cos
θ
)
(
1
−
cos
θ
)
\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}
(
1
+
c
o
s
θ
)
(
1
−
c
o
s
θ
)
(
1
+
s
i
n
θ
)
(
1
−
s
i
n
θ
)
,
(ii)
cot
2
θ
\cot ^{2} \theta
cot
2
θ
Q8
EXERCISE 8.1
If
3
cot
A
=
4
3 \cot \mathrm{A}=4
3
cot
A
=
4
, check whether
1
−
tan
2
A
1
+
tan
2
A
=
cos
2
A
−
sin
2
A
\frac{1-\tan ^{2} \mathrm{~A}}{1+\tan ^{2} \mathrm{~A}}=\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A}
1
+
t
a
n
2
A
1
−
t
a
n
2
A
=
cos
2
A
−
sin
2
A
or not.
Q9
EXERCISE 8.1
In triangle ABC , right-angled at B , if
tan
A
=
1
3
\tan \mathrm{A}=\frac{1}{\sqrt{3}}
tan
A
=
3
1
, find the value of:
(i)
sin
A
cos
C
+
cos
A
sin
C
\sin \mathrm{A} \cos \mathrm{C}+\cos \mathrm{A} \sin \mathrm{C}
sin
A
cos
C
+
cos
A
sin
C
(ii)
cos
A
cos
C
−
sin
A
sin
C
\cos \mathrm{A} \cos \mathrm{C}-\sin \mathrm{A} \sin \mathrm{C}
cos
A
cos
C
−
sin
A
sin
C
Q10
EXERCISE 8.1
In
△
P
Q
R
\triangle \mathrm{PQR}
△
PQR
, right-angled at
Q
,
P
R
+
Q
R
=
25
c
m
\mathrm{Q}, \mathrm{PR}+\mathrm{QR}=25 \mathrm{~cm}
Q
,
PR
+
QR
=
25
cm
and
P
Q
=
5
c
m
\mathrm{PQ}=5 \mathrm{~cm}
PQ
=
5
cm
. Determine the values of
sin
P
,
cos
P
\sin \mathrm{P}, \cos \mathrm{P}
sin
P
,
cos
P
and tan P .
Q11
EXERCISE 8.1
State whether the following are true or false. Justify your answer.
(i)
The value of
tan
A
\tan \mathrm{A}
tan
A
is always less than 1 .
(ii)
sec
A
=
12
5
\sec \mathrm{A}=\frac{12}{5}
sec
A
=
5
12
for some value of angle A .
(iii)
cos
A
\cos \mathrm{A}
cos
A
is the abbreviation used for the cosecant of angle A .
(iv)
cot
A
\cot \mathrm{A}
cot
A
is the product of
cot
\cot
cot
and A .
(v)
sin
θ
=
4
3
\sin \theta=\frac{4}{3}
sin
θ
=
3
4
for some angle
θ
\theta
θ
.
Q1
EXERCISE 8.2
Evaluate the following:
(i)
sin
60
∘
cos
30
∘
+
sin
30
∘
cos
60
∘
\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}
sin
6
0
∘
cos
3
0
∘
+
sin
3
0
∘
cos
6
0
∘
(ii)
2
tan
2
45
∘
+
cos
2
30
∘
−
sin
2
60
∘
2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}
2
tan
2
4
5
∘
+
cos
2
3
0
∘
−
sin
2
6
0
∘
(iii)
cos
45
∘
sec
30
∘
+
cosec
30
∘
\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}
s
e
c
3
0
∘
+
cosec
3
0
∘
c
o
s
4
5
∘
(iv)
sin
30
∘
+
tan
45
∘
−
cosec
60
∘
sec
30
∘
+
cos
60
∘
+
cot
45
∘
\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}
s
e
c
3
0
∘
+
c
o
s
6
0
∘
+
c
o
t
4
5
∘
s
i
n
3
0
∘
+
t
a
n
4
5
∘
−
cosec
6
0
∘
(v)
5
cos
2
60
∘
+
4
sec
2
30
∘
−
tan
2
45
∘
sin
2
30
∘
+
cos
2
30
∘
\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}
s
i
n
2
3
0
∘
+
c
o
s
2
3
0
∘
5
c
o
s
2
6
0
∘
+
4
s
e
c
2
3
0
∘
−
t
a
n
2
4
5
∘
Q2
EXERCISE 8.2
Choose the correct option and justify your choice :
(i)
2
tan
30
∘
1
+
tan
2
30
∘
=
\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=
1
+
t
a
n
2
3
0
∘
2
t
a
n
3
0
∘
=
(A)
sin
60
∘
\sin 60^{\circ}
sin
6
0
∘
(B)
cos
60
∘
\cos 60^{\circ}
cos
6
0
∘
(C)
tan
60
∘
\tan 60^{\circ}
tan
6
0
∘
(D)
sin
30
∘
\sin 30^{\circ}
sin
3
0
∘
(ii)
1
−
tan
2
45
∘
1
+
tan
2
45
∘
=
\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=
1
+
t
a
n
2
4
5
∘
1
−
t
a
n
2
4
5
∘
=
(A)
tan
90
∘
\tan 90^{\circ}
tan
9
0
∘
(B) 1
(C)
sin
45
∘
\sin 45^{\circ}
sin
4
5
∘
(D) 0
(iii)
sin
2
A
=
2
sin
A
\sin 2 \mathrm{~A}=2 \sin \mathrm{A}
sin
2
A
=
2
sin
A
is true when
A
=
\mathrm{A}=
A
=
(A)
0
∘
0^{\circ}
0
∘
(B)
30
∘
30^{\circ}
3
0
∘
(C)
45
∘
45^{\circ}
4
5
∘
(D)
60
∘
60^{\circ}
6
0
∘
(iv)
2
tan
30
∘
1
−
tan
2
30
∘
=
\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=
1
−
t
a
n
2
3
0
∘
2
t
a
n
3
0
∘
=
(A)
cos
60
∘
\cos 60^{\circ}
cos
6
0
∘
(B)
sin
60
∘
\sin 60^{\circ}
sin
6
0
∘
(C)
tan
60
∘
\tan 60^{\circ}
tan
6
0
∘
(D)
sin
30
∘
\sin 30^{\circ}
sin
3
0
∘
Q3
EXERCISE 8.2
If
tan
(
A
+
B
)
=
3
\tan (A+B)=\sqrt{3}
tan
(
A
+
B
)
=
3
and
tan
(
A
−
B
)
=
1
3
;
0
∘
B
\tan (A-B)=\frac{1}{\sqrt{3}} ; 0^{\circ}B
tan
(
A
−
B
)
=
3
1
;
0
∘
B
, find A and B.
Q4
EXERCISE 8.2
State whether the following are true or false. Justify your answer.
(i)
sin
(
A
+
B
)
=
sin
A
+
sin
B
\sin (A+B)=\sin A+\sin B
sin
(
A
+
B
)
=
sin
A
+
sin
B
.
(ii)
The value of
sin
θ
\sin \theta
sin
θ
increases as
θ
\theta
θ
increases.
(iii)
The value of
cos
θ
\cos \theta
cos
θ
increases as
θ
\theta
θ
increases.
(iv)
sin
θ
=
cos
θ
\sin \theta=\cos \theta
sin
θ
=
cos
θ
for all values of
θ
\theta
θ
.
(v)
cot
A
\cot \mathrm{A}
cot
A
is not defined for
A
=
0
∘
\mathrm{A}=0^{\circ}
A
=
0
∘
.
Q1
EXERCISE 8.3
Express the trigonometric ratios
sin
A
,
sec
A
\sin \mathrm{A}, \sec \mathrm{A}
sin
A
,
sec
A
and
tan
A
\tan \mathrm{A}
tan
A
in terms of
cot
A
\cot \mathrm{A}
cot
A
.
Q2
EXERCISE 8.3
Write all the other trigonometric ratios of
∠
A
\angle \mathrm{A}
∠
A
in terms of
sec
A
\sec \mathrm{A}
sec
A
.
Q3
EXERCISE 8.3
Choose the correct option. Justify your choice.
(i)
9
sec
2
A
−
9
tan
2
A
=
9 \sec ^{2} \mathrm{~A}-9 \tan ^{2} \mathrm{~A}=
9
sec
2
A
−
9
tan
2
A
=
(A) 1
(B) 9
(C) 8
(D) 0
(ii)
(
1
+
tan
θ
+
sec
θ
)
(
1
+
cot
θ
−
cosec
θ
)
=
(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=
(
1
+
tan
θ
+
sec
θ
)
(
1
+
cot
θ
−
cosec
θ
)
=
(A) 0
(B) 1
(C) 2
(D) -1
(iii)
(
sec
A
+
tan
A
)
(
1
−
sin
A
)
=
(\sec \mathrm{A}+\tan \mathrm{A})(1-\sin \mathrm{A})=
(
sec
A
+
tan
A
)
(
1
−
sin
A
)
=
(A)
sec
A
\sec \mathrm{A}
sec
A
(B)
sin
A
\sin \mathrm{A}
sin
A
(C)
cosec
A
\operatorname{cosec} \mathrm{A}
cosec
A
(D)
cos
A
\cos \mathrm{A}
cos
A
(iv)
1
+
tan
2
A
1
+
cot
2
A
=
\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}=
1
+
c
o
t
2
A
1
+
t
a
n
2
A
=
(A)
sec
2
A
\sec ^{2} \mathrm{~A}
sec
2
A
(B) -1
(C)
cot
2
A
\cot ^{2} A
cot
2
A
(D)
tan
2
A
\tan ^{2} \mathrm{~A}
tan
2
A
Q4
EXERCISE 8.3
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i)
(
cosec
θ
−
cot
θ
)
2
=
1
−
cos
θ
1
+
cos
θ
(\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}
(
cosec
θ
−
cot
θ
)
2
=
1
+
c
o
s
θ
1
−
c
o
s
θ
(ii)
cos
A
1
+
sin
A
+
1
+
sin
A
cos
A
=
2
sec
A
\frac{\cos \mathrm{A}}{1+\sin \mathrm{A}}+\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}}=2 \sec \mathrm{A}
1
+
s
i
n
A
c
o
s
A
+
c
o
s
A
1
+
s
i
n
A
=
2
sec
A
(iii)
tan
θ
1
−
cot
θ
+
cot
θ
1
−
tan
θ
=
1
+
sec
θ
cosec
θ
\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta
1
−
c
o
t
θ
t
a
n
θ
+
1
−
t
a
n
θ
c
o
t
θ
=
1
+
sec
θ
cosec
θ
(iv)
1
+
sec
A
sec
A
=
sin
2
A
1
−
cos
A
\frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{\sin ^{2} \mathrm{~A}}{1-\cos \mathrm{A}}
s
e
c
A
1
+
s
e
c
A
=
1
−
c
o
s
A
s
i
n
2
A
(v)
cos
A
−
sin
A
+
1
cos
A
+
sin
A
−
1
=
cosec
A
+
cot
A
\frac{\cos \mathrm{A}-\sin \mathrm{A}+1}{\cos \mathrm{A}+\sin \mathrm{A}-1}=\operatorname{cosec} \mathrm{A}+\cot \mathrm{A}
c
o
s
A
+
s
i
n
A
−
1
c
o
s
A
−
s
i
n
A
+
1
=
cosec
A
+
cot
A
, using the identity
cosec
2
A
=
1
+
cot
2
A
\operatorname{cosec}^{2} \mathrm{~A}=1+\cot ^{2} \mathrm{~A}
cosec
2
A
=
1
+
cot
2
A
.
(vi)
1
+
sin
A
1
−
sin
A
=
sec
A
+
tan
A
\sqrt{\frac{1+\sin \mathrm{A}}{1-\sin \mathrm{A}}}=\sec \mathrm{A}+\tan \mathrm{A}
1
−
s
i
n
A
1
+
s
i
n
A
=
sec
A
+
tan
A
(vii)
sin
θ
−
2
sin
3
θ
2
cos
3
θ
−
cos
θ
=
tan
θ
\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta
2
c
o
s
3
θ
−
c
o
s
θ
s
i
n
θ
−
2
s
i
n
3
θ
=
tan
θ
(viii)
(
sin
A
+
cosec
A
)
2
+
(
cos
A
+
sec
A
)
2
=
7
+
tan
2
A
+
cot
2
A
(\sin \mathrm{A}+\operatorname{cosec} \mathrm{A})^{2}+(\cos \mathrm{A}+\sec \mathrm{A})^{2}=7+\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}
(
sin
A
+
cosec
A
)
2
+
(
cos
A
+
sec
A
)
2
=
7
+
tan
2
A
+
cot
2
A
(ix)
(
cosec
A
−
sin
A
)
(
sec
A
−
cos
A
)
=
1
tan
A
+
cot
A
(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}
(
cosec
A
−
sin
A
)
(
sec
A
−
cos
A
)
=
t
a
n
A
+
c
o
t
A
1
(x)
(
1
+
tan
2
A
1
+
cot
2
A
)
=
(
1
−
tan
A
1
−
cot
A
)
2
=
tan
2
A
\left(\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}\right)=\left(\frac{1-\tan \mathrm{A}}{1-\cot \mathrm{A}}\right)^{2}=\tan ^{2} \mathrm{~A}
(
1
+
c
o
t
2
A
1
+
t
a
n
2
A
)
=
(
1
−
c
o
t
A
1
−
t
a
n
A
)
2
=
tan
2
A
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