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Byju's Answer
Standard XII
Mathematics
Trigonometric Ratios of Common Angles
The value of ...
Question
The value of
√
(
1
+
cos
θ
1
−
cos
θ
)
−
csc
θ
−
cot
θ
=
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Solution
√
(
1
+
cos
θ
)
(
1
−
cos
θ
)
×
(
1
+
cos
θ
)
(
1
+
cos
θ
)
−
cos
θ
−
cot
θ
=
1
+
cos
θ
sin
θ
−
c
o
s
e
c
θ
−
cot
θ
=
c
o
s
e
c
θ
+
cos
θ
−
c
o
s
e
s
θ
−
cot
θ
=
0
∴
The value is zero
(
0
)
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0
Similar questions
Q.
If
cot
θ
+
csc
θ
=
2
, then the value of
1
+
cos
θ
1
−
cos
θ
is
Q.
if
0
<
θ
<
90
∘
and
sin
θ
1
−
cos
θ
+
sin
θ
1
+
cos
θ
=
4
,
then the value of
θ
Q.
1
(
s
e
c
θ
−
t
a
n
θ
)
−
1
c
o
s
θ
=
1
c
o
s
θ
−
1
(
s
e
c
θ
+
t
a
n
θ
)
Q.
Prove
cot
θ
+
csc
θ
cot
θ
−
csc
θ
=
cos
θ
+
1
cos
θ
−
1
Q.
Whether the g
iven equation
cot
θ
−
cos
θ
cot
θ
+
cos
θ
=
csc
θ
−
1
csc
θ
+
1
?
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Standard XII Mathematics
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