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Byju's Answer
Standard X
Mathematics
Complementary Trigonometric Ratios
prove that se...
Question
prove that sec 17 / cosec 73 + tan 68 / cot 22 + cos square 44 + cos square 46
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Solution
sec
17
°
cosec
73
°
+
tan
68
°
cot
22
°
+
cos
2
44
°
+
cos
2
46
°
=
sec
90
°
-
73
°
cos
e
c
73
°
+
tan
90
°
-
22
°
cot
22
°
+
cos
2
90
°
-
46
°
+
cos
2
46
°
=
cos
e
c
73
°
cos
e
c
73
°
+
cot
22
°
cot
22
°
+
sin
2
46
°
+
cos
2
46
°
=
1
+
1
+
1
=
3
IDENTITY
USED
:
sec
90
°
-
θ
=
cosec
θ
tan
90
°
-
θ
=
cot
θ
cos
90
°
-
θ
=
sin
θ
sin
2
θ
+
cos
2
θ
=
1
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0
Similar questions
Q.
s
e
c
17
c
o
s
e
c
73
+
t
a
n
68
c
o
t
22
+
c
o
s
2
44
+
c
o
s
2
46
Q.
sec
17
∘
cosec
73
∘
+
tan
68
∘
cot
22
∘
−
[
cos
2
44
∘
+
cos
2
46
∘
]
=
___.
Q.
sec
17
∘
cosec
73
∘
+
tan
68
∘
cot
22
∘
−
[
cos
2
44
∘
+
cos
2
46
∘
]
=
___.
Q.
What is the answer of :
s
e
c
17
c
o
s
e
c
73
+
t
a
n
68
c
o
t
22
+
c
o
s
2
44
+
c
o
s
2
46
Q.
(i) Prove that
c
o
t
θ
+
c
o
s
e
c
θ
−
1
c
o
t
θ
−
c
o
s
e
c
θ
+
1
=
c
o
s
e
c
θ
+
c
o
t
θ
=
1
+
c
o
s
θ
s
i
n
θ
(ii) Prove that
1
s
e
c
A
−
t
a
n
A
−
1
c
o
s
A
=
1
c
o
s
A
−
1
s
e
c
A
+
t
a
n
A
.
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