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Mathematics
Mathematics
Problem
If
tan
y
=
2
t
1
−
t
2
\tan y = \frac{2t}{1-t^2}
tan
y
=
1
−
t
2
2
t
and
sin
x
=
2
t
1
+
t
2
\sin x = \frac{2t}{1+t^2}
sin
x
=
1
+
t
2
2
t
, then
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
(a)
2
1
+
t
2
\frac{2}{1+t^2}
1
+
t
2
2
(b)
1
1
+
t
2
\frac{1}{1+t^2}
1
+
t
2
1
(c) 1 (d) 2
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