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Mathematics
Mathematics
Problem
If
x
=
sin
−
1
(
2
t
1
+
t
2
)
x = \sin^{-1} \left( \frac{2t}{1+t^2} \right)
x
=
sin
−
1
(
1
+
t
2
2
t
)
and
y
=
tan
−
1
(
2
t
1
−
t
2
)
y = \tan^{-1} \left( \frac{2t}{1-t^2} \right)
y
=
tan
−
1
(
1
−
t
2
2
t
)
, then prove that
d
y
d
x
=
1
\frac{dy}{dx} = 1
d
x
d
y
=
1
.
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