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Mathematics
Problem
Let
f
(
x
)
=
(
sin
(
tan
−
1
x
)
+
sin
(
cot
−
1
x
)
)
2
−
1
f(x) = (\sin(\tan^{-1}x) + \sin(\cot^{-1}x))^2 - 1
f
(
x
)
=
(
sin
(
tan
−
1
x
)
+
sin
(
cot
−
1
x
)
)
2
−
1
,
∣
x
∣
>
1
|x| > 1
∣
x
∣
>
1
. If
d
y
d
x
=
1
2
d
d
x
(
sin
−
1
(
f
(
x
)
)
)
\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} (\sin^{-1}(f(x)))
d
x
d
y
=
2
1
d
x
d
(
sin
−
1
(
f
(
x
)))
and
y
(
3
)
=
π
6
y(\sqrt{3}) = \frac{\pi}{6}
y
(
3
)
=
6
π
, then
y
(
−
3
)
y(-\sqrt{3})
y
(
−
3
)
is equal to [JEE-MAIN ONLINE-2020] (1)
5
π
6
\frac{5\pi}{6}
6
5
π
(2)
−
π
6
-\frac{\pi}{6}
−
6
π
(3)
π
3
\frac{\pi}{3}
3
π
(4)
2
π
3
\frac{2\pi}{3}
3
2
π
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