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EAMCET Engineering 2007 Solved Paper
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© examsnet.com
Question : 107
Total: 160
s
i
n
A
+
s
i
n
B
=
√
3
(
cos
B
−
cos
A
)
⇒
s
i
n
3
A
+
s
i
n
3
B
is equal to
0
2
1
−
1
Validate
Solution:
Given,
s
i
n
A
+
s
i
n
B
=
√
3
(
cos
B
−
cos
A
)
⇒
s
i
n
A
+
√
3
cos
A
=
√
3
cos
B
−
s
i
n
B
⇒
1
2
s
i
n
A
+
√
3
2
cos
A
=
√
3
2
cos
B
−
1
2
s
i
n
B
⇒
cos
π
3
s
i
n
A
+
s
i
n
π
3
cos
A
=
s
i
n
π
3
cos
B
−
cos
π
3
s
i
n
B
⇒
s
i
n
(
A
+
π
3
)
=
s
i
n
(
π
3
−
B
)
⇒
A
+
π
3
=
π
3
−
B
⇒
A
=
−
B
Now,
s
i
n
3
(
A
)
+
s
i
n
3
B
=
s
i
n
(
−
3
B
)
+
s
i
n
3
B
=
−
s
i
n
3
B
+
s
i
n
3
B
=
0
© examsnet.com
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