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Byju's Answer
Standard XII
Mathematics
Derivative from First Principle
1+tan2x1-tan2...
Question
1
+
t
a
n
2
x
1
−
t
a
n
2
x
d
x
is equal to
A
l
o
g
1
−
t
a
n
x
1
+
t
a
n
x
+
c
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B
l
o
g
1
+
t
a
n
x
1
−
t
a
n
x
+
c
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C
1
2
l
o
g
1
−
t
a
n
x
1
+
t
a
n
x
+
c
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D
1
2
l
o
g
1
+
t
a
n
x
1
−
t
a
n
x
+
c
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Solution
The correct option is
D
1
2
l
o
g
1
+
t
a
n
x
1
−
t
a
n
x
+
c
∫
1
+
tan
2
(
x
)
1
−
tan
2
(
x
)
d
x
=
∫
sec
2
(
x
)
1
−
tan
2
(
x
)
d
x
A
p
p
l
y
u
−
s
u
b
s
t
i
t
u
t
i
o
n
:
u
=
tan
(
x
)
∫
1
1
−
u
2
d
u
U
s
e
t
h
e
c
o
m
m
o
n
i
n
t
e
g
r
a
l
:
∫
1
1
−
u
2
d
u
=
ln
|
u
+
1
|
2
−
ln
|
u
−
1
|
2
=
ln
|
u
+
1
|
2
−
ln
|
u
−
1
|
2
S
u
b
s
t
i
t
u
t
e
b
a
c
k
u
=
tan
(
x
)
=
ln
|
tan
(
x
)
+
1
|
2
−
ln
|
tan
(
x
)
−
1
|
2
A
d
d
a
c
o
n
s
t
a
n
t
t
o
t
h
e
s
o
l
u
t
i
o
n
=
ln
|
tan
(
x
)
+
1
|
2
−
ln
|
tan
(
x
)
−
1
|
2
+
C
=
1
2
l
o
g
1
+
t
a
n
x
1
−
t
a
n
x
+
C
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