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Byju's Answer
Standard IX
Mathematics
Use of Trigonometric Table
If 2cosθ+si...
Question
If
2
cos
θ
+
sin
θ
=
1
(
θ
≠
π
2
)
, then
7
cos
θ
+
6
sin
θ
is equal to:
A
11
2
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B
46
5
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C
1
2
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D
2
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Solution
The correct option is
D
2
2
cos
θ
+
sin
θ
=
1
θ
≠
π
2
2
cos
θ
=
1
−
sin
θ
4
cos
2
θ
=
(
1
−
sin
θ
)
2
Let
sin
θ
=
x
4
−
4
x
2
=
1
+
x
2
−
2
x
5
x
2
−
2
x
−
3
=
0
5
x
2
−
5
x
+
3
x
−
3
=
0
(
x
−
1
)
(
5
x
+
3
)
=
0
x
=
1
⇒
sin
θ
=
1
⇒
θ
=
π
2
but given
θ
≠
π
2
∴
x
=
−
3
5
sin
θ
=
−
3
5
∴
cos
θ
=
4
5
∴
7
cos
θ
+
6
sin
θ
=
7
×
4
5
+
6
×
−
3
5
=
10
5
=
2
Suggest Corrections
0
Similar questions
Q.
Assertion :If
2
cos
θ
+
sin
θ
=
1
(
θ
≠
π
2
)
then the value of
7
cos
θ
+
6
sin
θ
is
2
Reason: If
cos
2
θ
−
sin
θ
=
1
2
,
0
<
θ
<
π
2
, then
sin
θ
+
cos
6
θ
=
0
Q.
Assertion :If
tan
(
π
2
sin
θ
)
=
cot
(
π
2
cos
θ
)
, then
sin
(
θ
+
π
4
)
=
±
1
Reason:
−
√
2
≤
sin
(
θ
+
π
4
)
≤
√
2
Q.
If
sin
θ
−
√
6
cos
θ
=
√
7
cos
θ
. Prove that
cos
θ
−
√
6
sin
θ
−
√
7
sin
θ
=
1
.
Q.
Assertion :(A) : If
t
a
n
(
π
2
s
i
n
θ
)
=
c
o
t
(
π
2
c
o
s
θ
)
then
s
i
n
θ
+
c
o
s
θ
=
±
√
2
Reason: (R) :
−
√
2
≤
s
i
n
θ
+
c
o
s
θ
≤
√
2
Q.
IF
−
π
<
θ
<
−
π
2
, then
∣
∣
∣
√
1
−
sin
θ
1
+
sin
θ
+
√
1
+
sin
θ
1
−
sin
θ
∣
∣
∣
is equal to
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