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Mathematics
Problem
For
a
>
0
a > 0
a
>
0
,
t
∈
(
0
,
π
2
)
t \in \left(0, \frac{\pi}{2}\right)
t
∈
(
0
,
2
π
)
, let
x
=
a
sin
−
1
t
x = \sqrt{a^{\sin^{-1} t}}
x
=
a
s
i
n
−
1
t
and
y
=
a
cos
−
1
t
y = \sqrt{a^{\cos^{-1} t}}
y
=
a
c
o
s
−
1
t
. Then,
1
+
(
d
y
d
x
)
2
1 + \left(\frac{dy}{dx}\right)^2
1
+
(
d
x
d
y
)
2
equals : [JEE-MAIN ONLINE-2013] (1)
y
2
x
2
\frac{y^2}{x^2}
x
2
y
2
(2)
x
2
+
y
2
x
2
\frac{x^2 + y^2}{x^2}
x
2
x
2
+
y
2
(3)
x
2
y
2
\frac{x^2}{y^2}
y
2
x
2
(4)
x
2
+
y
2
y
2
\frac{x^2 + y^2}{y^2}
y
2
x
2
+
y
2
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