Prove that integration of
sec(y-x) =
[sec(y-x)-tan(y-x)]
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Answer:
Step-by-step explanation:
Given: y=tanx+secx
Prove that:
dx
2
d
2
y
=
(1−sinx)
2
cosx
y=
cosx
sinx
+
cosx
1
=
cosx
1+sinx
differentiate with respect to x
dx
dy
=
dx
d
(
cosx
1+sinx
)
dx
dy
=
cos
2
x
cosx
dx
d
(1+sinx)−(1+sinx)
dx
d
cosx
=
cos
2
x
cos
2
x+sinx+sin
2
x
=
cos
2
x
1+sinx
=
1−sin
2
x
1+sinx
=
(1+sinx)(1−sinx)
1+sinx
=
1−sinx
1
differentiate with respect to x
dx
d
(
dx
dy
)=
dx
d
(
1−sinx
1
)
dx
2
d
2
y
=
(1−sinx)
2
(1−sinx)
dx
d
(1)−(1)
dx
d
(1−sinx)
=
(1−sinx)
2
cosx
Hence proved.
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