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Standard XI
Mathematics
NCERT
Question
Prove that:
2
s
i
n
2
π
6
+
c
o
s
e
c
2
7
π
6
c
o
s
2
π
3
=
3
2
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Solution
Verified by Toppr
LHS
=
2
s
i
n
2
π
6
+
c
o
s
e
c
2
7
π
6
c
o
s
2
π
3
=
2
s
i
n
2
π
6
+
c
o
s
e
c
2
(
π
+
π
6
)
c
o
s
2
π
3
=
2
s
i
n
2
π
6
+
c
o
s
e
c
2
(
−
π
6
)
c
o
s
2
π
3
....As
θ
lies in the 3rd quadrant,
θ
is negative.
=
2
(
1
2
)
2
+
(
−
2
)
2
×
(
1
2
)
2
=
2
×
1
4
+
1
=
3
2
=
RHS
Hence proved.
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Similar Questions
Q1
Prove that:
2
s
i
n
2
π
6
+
c
o
s
e
c
2
7
π
6
c
o
s
2
π
3
=
3
2
View Solution
Q2
Prove that :
2
sin
2
π
6
+
cosec
2
7
π
6
cos
2
π
3
=
3
2
View Solution
Q3
Prove that
2
sin
2
π
6
+
cosec
2
7
π
6
cos
2
π
3
=
3
2
View Solution
Q4
The value of 2
×
2
sin
2
π
6
+
2
cosec
2
7
π
6
cos
2
π
3
View Solution
Q5
The value of
2
sin
2
π
6
+
cosec
2
7
π
6
⋅
cos
2
π
3
is
View Solution