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Byju's Answer
Standard XIII
Mathematics
Sin(A+B)Sin(A-B)
The value of ...
Question
The value of the expression
cos
20
∘
cos
100
∘
+
cos
100
∘
cos
140
∘
−
cos
140
∘
cos
200
∘
is
A
−
1
2
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B
−
3
4
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C
1
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D
−
1
2
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Solution
The correct option is
B
−
3
4
cos
20
∘
cos
100
∘
+
cos
100
∘
cos
140
∘
−
cos
140
∘
cos
200
∘
=
cos
100
∘
(
cos
20
∘
+
cos
140
∘
)
−
cos
140
∘
cos
200
∘
=
2
cos
100
∘
cos
80
∘
cos
60
∘
−
cos
140
∘
cos
200
∘
=
cos
100
∘
cos
80
∘
−
cos
140
∘
cos
200
∘
=
(
cos
180
∘
+
cos
20
∘
2
)
−
(
cos
340
∘
+
cos
60
∘
2
)
=
(
−
1
+
cos
20
∘
2
)
−
⎛
⎜ ⎜ ⎜
⎝
cos
20
∘
+
1
2
2
⎞
⎟ ⎟ ⎟
⎠
=
−
1
2
+
cos
20
∘
2
−
cos
20
∘
2
−
1
4
=
−
1
2
−
1
4
=
−
3
4
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0
Similar questions
Q.
Evaluate:
cos
20
∘
cos
100
∘
+
cos
100
∘
cos
140
∘
−
cos
140
∘
cos
200
∘
Q.
cos
20
∘
cos
100
∘
+
cos
100
∘
cos
140
∘
−
cos
140
∘
.
cos
200
∘
is equal to
Q.
Evaluate
cos
20
∘
+
cos
100
∘
+
cos
140
∘
.
Q.
Match the column
List I
List II
A.
cos
20
∘
+
cos
80
∘
−
√
3
cos
50
∘
1.
−
1
B.
cos
0
∘
+
cos
π
7
+
cos
2
π
7
+
cos
3
π
7
+
cos
4
π
7
+
cos
5
π
7
+
cos
6
π
7
2.
−
3
4
C.
cos
20
∘
+
cos
40
∘
+
cos
60
∘
+
4
cos
10
∘
cos
20
∘
cos
30
∘
3.
1
D.
cos
20
∘
cos
100
∘
+
cos
100
∘
cos
140
∘
−
cos
140
∘
cos
200
∘
4.
0
Q.
Prove that:
(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A
(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A
(iii) sin A + sin 2A + sin 4A + sin 5A = 4 cos
A
2
cos
3
A
2
sin 3A
(iv) sin 3A + sin 2A − sin A = 4 sin A cos
A
2
cos
3
A
2
(v) cos 20° cos 100° + cos 100° cos 140° − 140° cos 200° = −
3
4
(vi)
sin
θ
2
sin
7
θ
2
+
sin
3
θ
2
sin
11
θ
2
=
sin
2
θ
sin
5
θ
.
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