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Byju's Answer
Standard XII
Mathematics
Domain and Range of Basic Inverse Trigonometric Functions
Evaluate:- t...
Question
Evaluate:-
tan
−
1
2
{
cos
2
[
sin
−
1
(
1
2
)
]
}
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Solution
sin
−
1
1
2
=
θ
tan
−
1
2
(
cos
2
θ
)
tan
−
1
2
(
1
−
2
sin
2
θ
)
tan
−
1
2
(
1
−
2
(
(
sin
sin
−
1
1
2
)
)
2
)
tan
−
1
2
(
1
−
2
(
1
2
)
2
)
tan
−
1
2
(
1
−
1
2
)
tan
−
1
2
(
1
2
)
⇒
tan
−
1
=
1
=
45
∘
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0
Similar questions
Q.
Evaluate:
tan
−
1
(
1
)
+
cos
−
1
(
1
2
)
+
sin
−
1
(
1
2
)
which lies in the interval
[
0
,
π
]
Q.
Evaluate :
i)
tan
−
1
(
1
√
3
)
ii)
sin
−
1
(
−
1
2
)
Q.
Evaluate:
sin
39
∘
cos
51
∘
+
2
tan
11
∘
tan
31
∘
tan
45
∘
tan
59
∘
tan
79
∘
−
3
(
sin
2
21
∘
+
sin
2
69
∘
)
.
Q.
Evaluate each of the following:
(i)
cos
-
1
1
2
+
2
sin
-
1
1
2
(ii)
tan
-
1
2
cos
2
sin
-
1
1
2
(iii)
tan
-
1
1
+
cos
-
1
-
1
2
+
sin
-
1
-
1
2
(iv)
tan
-
1
3
-
sec
-
1
(
-
2
)
+
cosec
-
1
2
3
Q.
Evaluate each of the following:
i
tan
-
1
1
+
cos
-
1
-
1
2
+
sin
-
1
-
1
2
(ii)
tan
-
1
-
1
3
+
tan
-
1
-
3
+
tan
-
1
sin
-
π
2
(iii)
tan
-
1
tan
5
π
6
+
cos
-
1
cos
13
π
6
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