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Byju's Answer
Standard XII
Mathematics
Basic Inverse Trigonometric Functions
Values of c...
Question
Values of
cos
−
1
cos
13
π
2
+
tan
−
1
tan
13
π
5
+
sec
−
1
sec
13
π
3
+
cot
−
1
cot
17
π
2
+
sin
−
1
sin
33
π
5
is
A
23
π
5
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B
11
π
5
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C
4
π
3
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D
5
π
6
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Solution
The correct option is
C
4
π
3
cos
−
1
cos
13
π
2
=
π
2
tan
−
1
tan
13
π
5
=
13
π
5
−
3
π
sec
−
1
sec
13
π
3
=
13
π
3
−
4
π
cot
−
1
cot
17
π
2
=
π
2
sin
−
1
sin
33
π
5
=
7
π
−
33
π
5
∴
Sum
=
π
2
+
π
2
+
π
3
=
4
π
3
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Similar questions
Q.
Prove that:
(i)
sin
A
+
sin
3
A
+
sin
5
A
cos
A
+
cos
3
A
+
cos
5
A
=
tan
3
A
(ii)
cos
3
A
+
2
cos
5
A
+
cos
7
A
cos
A
+
2
cos
3
A
+
cos
5
A
=
cos
5
A
cos
3
A
(iii)
cos
4
A
+
cos
3
A
+
cos
2
A
sin
4
A
+
sin
3
A
+
sin
2
A
=
cot
3
A
(iv)
sin
3
A
+
sin
5
A
+
sin
7
A
+
sin
9
A
cos
3
A
+
cos
5
A
+
cos
7
A
+
cos
9
A
=
tan
6
A
(v)
sin
5
A
-
sin
7
A
+
sin
8
A
-
sin
4
A
cos
4
A
+
cos
7
A
-
cos
5
A
-
cos
8
A
=
cot
6
A
(vi)
sin
5
A
cos
2
A
-
sin
6
A
cos
A
sin
A
sin
2
A
-
cos
2
A
cos
3
A
=
tan
A
(vii)
sin
11
A
sin
A
+
sin
7
A
sin
3
A
cos
11
A
sin
A
+
cos
7
A
sin
3
A
=
tan
8
A
(viii)
sin
3
A
cos
4
A
-
sin
A
cos
2
A
sin
4
A
sin
A
+
cos
6
A
cos
A
=
tan
2
A
(ix)
sin
A
sin
2
A
+
sin
3
A
sin
6
A
sin
A
cos
2
A
+
sin
3
A
cos
6
A
=
tan
5
A
(x)
sin
A
+
2
sin
3
A
+
sin
5
A
sin
3
A
+
2
sin
5
A
+
sin
7
A
=
sin
3
A
sin
5
A
(xi)
sin
θ
+
ϕ
-
2
sin
θ
+
sin
θ
-
ϕ
cos
θ
+
ϕ
-
2
cos
θ
+
cos
θ
-
ϕ
=
tan
θ
Q.
If
5
tan
θ
=
4
, then value of
5
sin
θ
−
3
cos
θ
5
sin
θ
+
2
cos
θ
is:
Q.
Prove that:
(i)
tan
4
π
-
cos
3
π
2
-
sin
5
π
6
cos
2
π
3
=
1
4
(ii)
sin
13
π
3
sin
8
π
3
+
cos
2
π
3
sin
5
π
6
=
1
2
(iii)
sin
13
π
3
sin
2
π
3
+
cos
4
π
3
sin
13
π
6
=
1
2
(iv)
sin
10
π
3
cos
13
π
6
+
cos
8
π
3
sin
5
π
6
=
-
1
(v)
tan
5
π
4
cot
9
π
4
+
tan
17
π
4
cot
15
π
4
=
0
Q.
The value of
tan
[
2
sin
−
1
3
5
+
cos
−
1
3
5
]
is
Q.
If
5
tan
α
= 4, show that
5
s
i
n
α
−
3
c
o
s
α
5
s
i
n
α
+
2
c
o
s
α
=
1
6
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