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TS EAMCET 4 May 2018 Shift 1 Solved Paper
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© examsnet.com
Question : 21
Total: 160
Whwn
s
i
n
9
θ
c
o
s
27
θ
+
s
i
n
3
θ
c
o
s
9
θ
+
s
i
n
θ
c
o
s
3
θ
= k (tan 27 θ - tan θ) is defined , then k =
π
2
−
1
2
1
2
π
4
Validate
Solution:
Consider the trigonometric function.
sin
9
θ
cos
27
θ
+
sin
3
θ
cos
9
θ
+
sin
θ
cos
3
θ
=
k
(
tan
27
θ
−
tan
θ
)
Consider the left hand side of the function.
LHS
=
sin
9
θ
cos
27
θ
+
sin
3
θ
cos
9
θ
+
sin
θ
cos
3
θ
=
sin
9
θ
cos
9
θ
+
sin
3
θ
cos
27
θ
cos
9
θ
cos
27
θ
+
sin
θ
cos
3
θ
=
sin
18
θ
+
sin
30
θ
−
sin
24
θ
2
cos
9
θ
cos
27
θ
+
sin
θ
cos
3
θ
=
[
(
sin
21
θ
+
sin
15
θ
+
sin
33
θ
+
sin
27
θ
−
sin
27
θ
−
sin
21
θ
+
4
sin
θ
cos
9
θ
cos
27
θ
)
4
cos
3
θ
cos
9
θ
cos
27
θ
]
Solve further,
LHS
=
[
sin
15
θ
+
sin
33
θ
+
4
sin
θ
cos
9
θ
cos
27
θ
4
cos
3
θ
cos
9
θ
cos
27
θ
]
=
[
2
sin
24
θ
cos
9
θ
+
4
sin
θ
cos
9
θ
cos
27
θ
4
cos
3
θ
cos
9
θ
cos
27
θ
]
=
[
sin
24
θ
+
2
sin
θ
cos
27
θ
2
cos
3
θ
cos
27
θ
]
=
[
sin
(
27
θ
−
3
θ
)
2
cos
3
θ
cos
27
θ
+
sin
θ
cos
3
θ
]
Solve further,
LHS
=
1
2
[
sin
27
θ
cos
27
θ
−
sin
3
θ
cos
3
θ
]
+
sin
θ
cos
3
θ
=
1
2
sin
27
θ
cos
27
θ
−
1
2
(
sin
3
θ
−
2
sin
θ
cos
3
θ
)
=
1
2
tan
27
θ
−
1
2
(
3
sin
θ
−
4
sin
3
θ
−
2
sin
θ
cos
θ
(
4
cos
2
θ
−
3
)
)
=
1
2
[
tan
27
θ
−
sin
θ
(
1
−
4
sin
2
θ
)
cos
θ
(
1
−
4
sin
2
θ
)
]
Solve further,
LHS
=
1
2
[
tan
27
θ
−
tan
θ
]
Compare with RHS.
k
=
1
2
© examsnet.com
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