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Standard XII
Mathematics
Cos(A+B)Cos(A-B)
Question
Prove that :
c
o
s
(
π
6
−
A
)
⋅
c
o
s
(
π
3
+
B
)
−
s
i
n
(
π
6
−
A
)
⋅
s
i
n
(
π
3
+
B
)
=
cos
(
A
−
B
)
Open in App
Solution
Verified by Toppr
cos
(
π
6
−
A
)
cos
(
π
3
+
B
)
−
sin
(
π
6
−
A
)
sin
(
π
3
+
B
)
Using compound angle formula,
cos
(
A
+
B
)
=
cos
A
cos
B
−
sin
A
sin
B
=
cos
(
π
6
−
A
+
π
3
+
B
)
=
cos
(
π
+
2
π
6
−
A
+
B
)
=
cos
(
3
π
6
−
A
+
B
)
=
cos
(
π
2
−
(
A
−
B
)
)
=
cos
(
A
−
B
)
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3
Similar Questions
Q1
Prove that :
c
o
s
(
π
6
−
A
)
⋅
c
o
s
(
π
3
+
B
)
−
s
i
n
(
π
6
−
A
)
⋅
s
i
n
(
π
3
+
B
)
=
cos
(
A
−
B
)
View Solution
Q2
Prove the following:
sin
(
π
6
+
A
)
⋅
cos
(
π
3
−
B
)
+
sin
(
π
3
−
B
)
⋅
cos
(
π
6
+
A
)
=
cos
A
−
B
View Solution
Q3
Find
cos
(
π
6
−
A
)
cos
(
π
3
+
B
)
−
sin
(
π
6
−
A
)
sin
(
π
3
+
B
)
=
.
View Solution
Q4
Prove that:
sin
[
π
6
+
A
]
.
cos
[
π
3
−
B
]
+
sin
[
π
3
−
B
]
.
cos
[
π
6
+
A
]
=
cos
(
A
−
B
)
View Solution
Q5
What is
⎡
⎢ ⎢ ⎢ ⎢
⎣
sin
π
6
+
i
(
1
−
cos
π
6
)
sin
π
6
−
i
(
1
−
cos
π
6
)
⎤
⎥ ⎥ ⎥ ⎥
⎦
3
where
i
=
√
−
1
, equal to?
View Solution