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JEE Main Math Class 12 Inverse Trigonometric Functions Questions
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© examsnet.com
Question : 9
Total: 74
If
S
=
{
x
∈
R
:
s
i
n
−
1
(
x
+
1
√
x
2
+
2
x
+
2
)
−
s
i
n
−
1
(
x
√
x
2
+
1
)
=
π
4
}
, then is equal to _______.
[13-Apr-2023 shift 1]
Your Answer:
Validate
Solution:
s
i
n
−
1
(
(
x
+
1
)
√
(
x
+
1
)
2
+
1
)
−
s
i
n
−
1
(
x
√
x
2
+
1
)
=
π
4
∵
t
√
t
2
+
1
∈
(
−
1
,
1
)
s
i
n
−
1
(
(
x
+
1
)
√
(
x
+
1
)
2
+
1
)
=
s
i
n
−
1
(
x
√
x
2
+
1
)
+
π
4
(
x
+
1
)
√
(
x
+
1
)
2
+
1
=
(
1
√
2
)
cos
(
s
i
n
−
1
(
x
√
x
2
+
1
)
)
+
1
√
2
(
x
√
x
2
+
1
)
=
1
√
2
(
1
√
x
2
+
1
+
x
√
x
2
+
1
)
(
x
+
1
)
√
(
x
+
1
)
2
+
1
=
1
√
2
(
1
+
x
√
x
2
+
1
)
After solving this equation, we get
x
=
−
1
or
x
=
0
S
=
{
−
1
,
0
}
∑
x
∈
R
(
s
i
n
(
(
x
2
+
x
+
5
)
π
2
)
−
cos
(
(
x
2
+
x
+
5
)
π
)
)
=
[
s
i
n
(
5
π
2
)
−
cos
(
5
π
)
]
+
[
s
i
n
(
5
π
2
)
−
cos
(
5
π
)
]
=
4
© examsnet.com
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