Exercise 10.2 (Solutions)
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Textbook of Algebra and
and Trigonometry for Class XI
Merging man and maths
Available online @ http://www.mathcity.org, Version: 2.0.2
Question # 1
(i) L.H.S = sin(180 + θ ) = sin180 cosθ + cos180 sin θ
= sin ( 0 ) cosθ + (−1)sin θ = 0 − sin θ = − sin θ = R.H.S (ii) Do youself
(iii)
tan 270 − tan θ
L.H.S = tan ( 270 − θ ) =
1 + tan 270 tan θ
tan θ 

 tan θ 
tan 270 1 −
1 −


tan 270 
∞ 


=
=
1
 1


tan 270 
+ tan θ   + tan θ 
 tan 270
 ∞

(1 − 0 ) = 1 = cotθ = R.H.S =
( 0 + tanθ ) tan θ
Remaining do yourself.
Question # 2
(i) Since 15 = 45 – 30
So sin15 = sin(45 − 30) = sin 45 cos30 − cos 45 sin 30
3
1
3 −1
 1   3   1  1 
=
−
=
 2  − 
 2  =
2
2
2
2
2
2
2


 2 



(ii) Since 15 = 45 – 30
So cos15 = cos(45 − 30) = cos 45 cos30 + sin 45 sin 30
3
1
3 +1
 1   3   1  1 
=
+
=
+
=







2 2
 2  2   2  2  2 2 2 2
(iii) Since 15 = 45 – 30
tan 45 − tan 30
So tan15 = tan(45 − 30) =
1 + tan 45 tan 30
1
1
1−
1−
3
3
=
=
1
 1 
1+
1 + (1) 

3
 3
=
3 −1
3
3 +1
3
=
3 −1
3 +1
FSc-I / 10.2 - 2
1
1
=
cos15 cos(45 − 30)
Now do yourself
(v), (vi) and (vii)
Hint: Use 105 = 60 + 45 in these questions
(iv) sec15 =
Question # 3
(i) L.H.S = sin(45 + α )
1
 1

= sin 45 cosα + cos 45 sin α = 
cosα +
sin α ) 
2
 2

1
1
=
( cosα + sin α ) = ( sin α + cosα ) = R.H.S 2
2
(ii)
Do youself as above
Question # 4 (i)
L.H.S = tan(45 + A) tan(45 − A)
 tan 45 + tan A   tan 45 − tan A 
=


 1 − tan 45 tan A   1 + tan 45 tan A 
 1 + tan A   1 − tan A 
 1 + tan A   1 − tan A 
=

 =  1 − tan A   1 + tan A  = 1 = R.H.S 


 1 − (1) tan A   1 + (1) tan A 
Question 4 (ii)
π

 3π

L.H.S = tan  − θ  + tan 
+θ 
4

 4

π
3π

 
tan
−
tan
θ
tan
+ tan θ

 
4
4
=
+
π
3π
 1 + tan tan θ   1 − tan
tan θ

4
 
4
 1 − tan θ   −1 + tan θ 
=
+

1
+
(1)
tan
θ

  1 − (−1) tan θ 





 1 − tan θ   −1 + tan θ 
=
+

1
+
tan
θ

  1 + tan θ 
1 − tan θ − 1 + tan θ
0
=
=
= 0 = R.H.S 1 + tan θ
1 + tan θ
Question # 4 (iii)
π
π


L.H.S = sin θ +  + cos θ + 
6
3


FSc-I / 10.2 - 3
π
π 
π
π

=  sin θ cos + cosθ sin  +  cosθ cos − sin θ sin 
6
6 
3
3


3
1 
1
3
=  sin θ
+ cosθ  +  cosθ − sin θ

2
2 
2
2 

3
1
1
3
sin θ + cosθ + cosθ −
sin θ = cosθ = R.H.S 2
2
2
2
=
Question # 4 (iv)
sin θ − cosθ tan
L.H.S =
cosθ + sin θ tan
θ
2
θ
sin
sin θ − cosθ
cos
=
sin
cosθ + sin θ
cos
2
θ
sin θ cos
2
θ
2
θ
2
θ
2
− cosθ sin
cos
=
cosθ cos
θ
θ
2
2
+ sin θ sin
cos
2
θ
θ
θ
2
θ
2
2
θ

θ
θ
sin θ − 
sin θ cos − cosθ sin
sin θ
2
2

2
2
=
=
=
θ
θ
θ

cos θ
cosθ cos + sin θ sin
cos θ − 
2
2
2
2

( )
( )
= tan
θ
2
= R.H.S Question # 4 (v)
1 − tan θ tan ϕ
L.H.S =
1 + tan θ tan ϕ
sin θ sin ϕ
cosθ cos ϕ − sin θ sin ϕ
1−
cosθ cos ϕ
cosθ cos ϕ
=
=
sin θ sin ϕ
cosθ cos ϕ + sin θ sin ϕ
1+
cosθ cos ϕ
cosθ cos ϕ
cos (θ + ϕ )
cosθ cos ϕ − sin θ sin ϕ
=
=
= R.H.S cosθ cos ϕ + sin θ sin ϕ
cos (θ − ϕ )
Question # 5
L.H.S = cos(α + β ) cos(α − β )
= ( cosα cos β − sin α sin β )( cosα cos β + sin α sin β )
(
2
= ( cosα cos β ) − ( sin α sin β )
2
) = cos α cos β − sin α sin
2
= cos 2 α (1 − sin 2 β ) − (1 − cos 2 α ) sin 2 β
2
2
2
β
FSc-I / 10.2 - 4
= cos 2 α − cos 2 α sin 2 β − sin 2 β + cos 2 α sin 2 β
= cos 2 α − sin 2 β …………….. (i)
= (1 − sin 2 α ) − (1 − cos 2 β )
= 1 − sin 2 α − 1 + cos 2 β
= cos 2 β − sin 2 α …………….. (ii)
Question # 6
Hint: Just open the formulas
Question # 7 (i)
L.H.S = cot(α + β ) =
1
tan(α + β )
=
1
tan α + tan β
1 − tan α tan β


1
tan α tan β 
− 1
1 − tan α tan β
 tan α tan β

=
=
tan α + tan β
 1
1 
+
tan α tan β 

 tan β tan α 
cot α cot β − 1
cot α cot β − 1
=
=
= R.H.S cot β + cot α
cot α + cot β
Question # 7 (ii)
Do yourself as above
Question # 7 (iii)
sin α sin β
sin α cos β + cosα sin β
+
tan α + tan β
cosα cos β
cosα cos β
L.H.S =
=
=
sin α sin β
sin α cos β − cosα sin β
tan α − tan β
−
cosα cos β
cosα cos β
sin α cos β + cosα sin β
sin(α + β )
=
=
= R.H.S
sin α cos β − cosα sin β
sin(α − β )
Question # 8
4
π
Since sin α =
;
0 <α <
2
5
40
π
cosα =
;
0< β <
41
2
Now
cos 2 α = 1 − sin 2 α
⇒ cosα = ± 1 − sin 2 α
Since terminal ray of α is in the first quadrant so value of cos is +ive
FSc-I / 10.2 - 5
cosα = 1 − sin 2 α
4
⇒ cosα = 1 −  
5
2
= 1−
16
9
=
25
25
⇒ cosα =
3
5
Also
sin 2 β = 1 − cos 2 β
⇒ sin β = ± 1 − cos 2 β
Since terminal ray of β is in the first quadrant so value of sin is +ive
sin β = 1 − cos 2 β
2
1600
81
 40 
⇒ sin β = 1 −   = 1 −
=
1681
1681
 41 
⇒ sin β =
9
41
Now
sin(α − β ) = sin α cos β − cosα sin β
160 27
133
 4  40   3  9 
=    −    =
−
=
205 205 205
 5  41   5  41 
133
i.e. sin(α − β ) =
Proved
205
Question # 9
4
Since sin α =
;
5
12
sin β =
;
13
π
2
π
2
<α <π
< β <π
Since
cos 2 α = 1 − sin 2 α ⇒ cosα = ± 1 − sin 2 α
As terminal ray of α lies in the IInd quadrant so value of cos is –ive
cosα = − 1 − sin 2 α
2
16
9
4
⇒ cosα = − 1 −   = − 1 −
=−
25
25
5
⇒ cosα = −
3
5
Now
cos 2 β = 1 − sin 2 β
⇒ cos β = ± 1 − sin 2 β
As terminal ray of β lies in the IInd quadrant so value of cos is –ive
cos β = − 1 − sin 2 β
2
(i)
144
25
5
 12 
⇒ cos β = − 1 −   = − 1 −
=−
⇒ cos β = −
169
169
13
 13 
sin(α + β ) = sin α cos β + cosα sin β
20 36
56
 4  5   3  12 
=   −  +  −   = − −
=−
65 65
65
 5  13   5  13 
FSc-I / 10.2 - 6
cos(α + β ) = cosα cos β − sin α sin β
33
 3  5   4  12  15 48
=  −  −  −    =
−
=−
65
 5  13   5  13  65 65
56
sin(α + β ) − 65 56
(iii) tan(α + β ) =
=
=
cos(α + β ) − 33
33
65
(iv), (v) & (vi) Do yourself as above
Since sin(α + β ) is –ive so terminal are of α + β is in IIIrd or IVth quadrant
and cos(α + β ) is –ive so terminal are of α + β is in IInd or IIIrd quadrant
therefore terminal ray of α + β lies in the IIIrd quadrant.
Similarly after solving (iv), (v) & (vi) find quadrant for α − β yourself.
(ii)
Question # 10 (i)
3
Since tan α =
4
tanα is +ive and terminal arm of α in not in the Ist quad. Therefore it lies in
IIIrd quad.
Now
sec2 α = 1 + tan 2 α
⇒ secα = ± 1 + tan 2 α
Since terminal arm of α is in the IIIrd quad. therefore value of sec is –ive
secα = − 1 + tan 2 α
2
9
25
5
3
⇒ secα = − 1 +   = − 1 +
=−
⇒ sec x = −
4
16
16
4
1
1
4
Now
cosα =
=
⇒ cosα = −
secα − 5
5
4
sin α
Now
= tan α
⇒ sin α = tan α cosα
cosα
3
 3  4 
⇒ sin α =   − 
⇒ sin α = −
5
 4  5 
5
Since cos β =
13
cos β is +ive and terminal arm of β is not in the Ist quad. Therefore it lies in
IVth quad.
sin 2 β = 1 − cos 2 β
Now
⇒ sin β = ± 1 − cos 2 β
Since terminal ray of β is in the fourth quadrant so value of sin is –ive
sin β = − 1 − cos 2 β
FSc-I / 10.2 - 7
2
25
144
5
⇒ sin β = − 1 −   = − 1 −
=−
169
169
 13 
⇒ sin β = −
12
13
Now sin(α + β ) = sin α cos β + cosα sin β
3 48
 3   5   4   12 
= −    + −  −  = − +
13 65
 5   13   5   13 
⇒ sin(α + β ) =
33
65
And cos(α + β ) = cosα cos β − sin α sin β
4 36
 4  5   3  12 
=  −   −  −  −  = − −
13 65
 5  13   5  13 
⇒ cos(α + β ) = −
Question # 10 (ii)
Do yourself as above
Question # 11
R.H.S = tan 37 = tan(45 − 8)
∵ 37 = 45 − 8
tan 45 − tan8
1 − tan8
=
1 + (1) tan8
1 + tan 45 tan8
sin8
cos8 − sin8
1−
cos8 − sin8
cos8
cos8
=
=
= L.H.S =
sin8
cos8 + sin8
cos8 + sin8
1+
cos8
cos8
=
Question # 12
Since α , β and γ are angles of triangle therefore
α + β + γ = 180 ⇒ α + β = 180 − γ
α + β 180 − γ
α β
γ
⇒
=
⇒ + = 90 −
2
2
2 2
2
γ
α β 

tan  +  = tan  90 − 
Now
2
2 2

⇒
⇒
tan
α
+ tan
β
2 = cot γ
α
β
2
1 − tan tan
2
2
α
β 1
1
tan tan 
+
2
2  tan β2 tan α2
tan
2
α
2
tan
β
1

2  tan α2 tan β2


 = cot γ

2
− 1

γ
γ

∵ tan  90 −  = cot
2
2

⇒
cot
β
+ cot
α
2 = cot γ
α
β
2
cot cot − 1
2
2
2
56
65
FSc-I / 10.2 - 8
⇒ cot
β
⇒ cot
β
⇒ cot
β
2
2
2
+ cot
+ cot
+ cot
γ α
β 
= cot  cot cot − 1
2
2
2
2

α
α
2
α
2
= cot
+ cot
α
2
γ
2
cot
β
2
= cot
cot
α
2
γ
2
cot
− cot
β
2
cot
γ
2
γ
2
Question # 13
Since α , β and γ are angles of triangle therefore
α + β + γ = 180 ⇒ α + β = 180 − γ
Now tan (α + β ) = tan (180 − γ )
tan α + tan β
⇒
= tan ( 2(90) − γ )
1 − tan α tan β
tan α + tan β
⇒
= − tan γ
1 − tan α tan β
⇒ tan α + tan β = − tan γ (1 − tan α tan β )
⇒ tan α + tan β = − tan γ + tan α tan β tan γ
⇒ tan α + tan β + tan γ = tan α tan β tan γ
Dividing through out by tan α tan β tan γ
tan α
tan β
tan γ
tan α tan β tan γ
+
+
=
tan α tan β tan γ tan α tan β tan γ tan α tan β tan γ tan α tan β tan γ
⇒ cot β cot γ + cot α cot γ + cot α cot β = 1
⇒ cot α cot β + cot β cot γ + cot γ cot α = 1
Question # 14 (i)
12sin θ + 5cosθ
Let 12 = r cos ϕ and 5 = r sin ϕ
Squaring and adding
(12) 2 + (5) 2 = r 2 cos 2 ϕ + r 2 sin 2 ϕ
⇒ 144 + 25 = r 2 ( cos 2 ϕ + sin 2 ϕ )
⇒ 169 = r 2 (1)
⇒ r = 169 = 13
Now
5 r sin ϕ
=
12 r cos ϕ
5
= tan ϕ
12
5
⇒ ϕ = tan −1
12
12sin θ + 5cosθ = r cos ϕ sin θ + r sin ϕ cosθ
= r ( cos ϕ sin θ + sin ϕ cosθ )
= r sin (θ + ϕ )
where r = 13 and ϕ = tan −1
5
12
FSc-I / 10.2 - 9
Question # 14 (ii)
3sin θ − 4cosθ
Let 3 = r cos ϕ
and −4 = r sin ϕ
Squaring and adding
(3) 2 + (−4) 2 = r 2 cos 2 ϕ + r 2 sin 2 ϕ
− 4 r sin ϕ
=
3 r cos ϕ
4
− = tan ϕ
3
 4
⇒ ϕ = tan −1  − 
 3
⇒ 9 + 16 = r 2 ( cos 2 ϕ + sin 2 ϕ )
⇒ 25 = r 2 (1)
⇒ r = 25 = 5
Now
3sin θ − 4cosθ = r cos ϕ sin θ + r sin ϕ cosθ
= r ( cos ϕ sin θ + sin ϕ cosθ )
= r sin (θ + ϕ )
 4
where r = 5 and ϕ = tan −1  − 
 3
Made by: Atiq ur Rehman ([email protected]) http://www.mathcity.org
Question # 14 (iii)
sin θ − cosθ
Let 1 = r cos ϕ
and −1 = r sin ϕ
Squaring and adding
(1) 2 + (−1)2 = r 2 cos 2 ϕ + r 2 sin 2 ϕ
−1 r sin ϕ
=
1 r cos ϕ
−1 = tan ϕ
⇒ ϕ = tan −1 ( −1)
⇒ 1 + 1 = r 2 ( cos 2 ϕ + sin 2 ϕ )
⇒ 2 = r 2 (1)
⇒ r= 2
Now
sin θ − cosθ = r cos ϕ sin θ + r sin ϕ cosθ
= r ( cos ϕ sin θ + sin ϕ cosθ )
= r sin (θ + ϕ )
Question # 14 (iv)
5sin θ − 4cosθ
Let 5 = r cos ϕ
and −4 = r sin ϕ
Squaring and adding
(5) 2 + (−4) 2 = r 2 cos 2 ϕ + r 2 sin 2 ϕ
⇒ 25 + 16 = r 2 ( cos 2 ϕ + sin 2 ϕ )
⇒ 41 = r 2 (1)
⇒ r = 41
where r = 2 and ϕ = tan −1 ( −1)
− 4 r sin ϕ
=
5 r cos ϕ
4
− = tan ϕ
5
 4
⇒ ϕ = tan −1  − 
 5
FSc-I / 10.2 - 10
Now
5sin θ − 4cosθ = r cos ϕ sin θ + r sin ϕ cosθ
= r ( cos ϕ sin θ + sin ϕ cosθ )
= r sin (θ + ϕ )
 4
where r = 41 and ϕ = tan −1  − 
 5
Question # 14 (v)
sin θ + cosθ
and 1 = r sin ϕ
Let 1 = r cos ϕ
Squaring and adding
(1) 2 + (1) 2 = r 2 cos 2 ϕ + r 2 sin 2 ϕ
1 r sin ϕ
=
1 r cos ϕ
1 = tan ϕ
⇒ ϕ = tan −1 (1)
⇒ 1 + 1 = r 2 ( cos 2 ϕ + sin 2 ϕ )
⇒ 2 = r 2 (1)
⇒ r= 2
Now
sin θ + cosθ = r cos ϕ sin θ + r sin ϕ cosθ
= r ( cos ϕ sin θ + sin ϕ cosθ )
= r sin (θ + ϕ )
where r = 2 and ϕ = tan −1 (1)
Question # 14 (vi)
Do yourself as above ☺
Book:
Exercise 10.2
Text Book of Algebra and Trigonometry Class XI
Punjab Textbook Board, Lahore.
Made by: Atiq ur Rehman ([email protected])
Available online at http://www.MathCity.org in PDF Format
(Picture format to view online).
Page setup: A4 (8.27 in × 11.02 in).
Updated: 29-12-2014.