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Inverse Trigonometric Functions
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Section:
Mathematics
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© examsnet.com
Question : 38
Total: 89
If
0
<
x
<
‌
1
√
2
and
‌
sin
−
1
x
α
=
‌
cos
−
1
x
β
, then the value of
sin
(
‌
2
Ï€
α
α
+
β
)
is
[26-Jul-2022-Shift-2]
4
√
(
1
−
x
2
)
(
1
−
2
x
2
)
4
x
√
(
1
−
x
2
)
(
1
−
2
x
2
)
2
x
√
(
1
−
x
2
)
(
1
−
4
x
2
)
4
√
(
1
−
x
2
)
(
1
−
4
x
2
)
Validate
Solution:
👈: Video Solution
Let
‌
sin
−
1
x
α
=
‌
cos
−
1
x
β
=
k
⇒
sin
−
1
x
+
cos
−
1
x
=
k
(
α
+
β
)
⇒
α
+
β
=
‌
Ï€
2
k
Now,
‌
2
Ï€
α
α
+
β
=
‌
2
Ï€
α
Ï€
2
k
=
4
k
α
=
4
sin
−
1
x
Here
sin
(
‌
2
Ï€
α
α
+
β
)
=
sin
(
4
sin
−
1
x
)
Let
sin
−
1
x
=
θ
∵
x
∈
(
0
,
‌
1
√
2
)
⇒
θ
∈
(
0
,
‌
Ï€
4
)
⇒
x
=
sin
‌
θ
⇒
cos
‌
θ
=
√
1
−
x
2
⇒
sin
‌
2
‌
θ
=
2
x
â‹…
√
1
−
x
2
⇒
cos
‌
2
‌
θ
=
√
1
−
4
x
2
(
1
−
x
2
)
=
√
(
2
x
2
−
1
)
2
=
1
−
2
x
2
∵
(
cos
‌
2
‌
θ
>
0.
as
.2
θ
∈
(
0
,
‌
Ï€
2
)
)
⇒
sin
‌
4
‌
θ
=
2
â‹…
2
x
√
1
−
x
2
(
1
−
2
x
2
)
=
4
x
√
1
−
x
2
(
1
−
2
x
2
)
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