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Byju's Answer
Standard IX
History
Machiavelli
If α =∫01e9x+...
Question
If
α
=
1
∫
0
(
e
9
x
+
3
tan
−
1
x
)
(
12
+
9
x
2
1
+
x
2
)
d
x
where
tan
–
1
x
takes only principal values, then the value of
(
log
e
|
1
+
α
|
−
3
π
4
)
is
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Solution
α
=
∫
1
0
(
e
9
x
+
3
tan
1
x
)
(
12
+
9
x
2
1
+
x
2
)
d
x
=
∫
1
0
(
e
9
x
+
3
tan
1
x
)
(
9
+
3
1
+
x
2
)
d
x
=
(
e
9
x
+
3
tan
1
x
)
∣
∣
1
0
=
e
9
+
3
π
4
−
1
⇒
ln
|
(
1
+
α
)
|
−
3
π
4
=
9
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0
Similar questions
Q.
If
α
=
∫
1
0
(
e
(
9
x
+
3
tan
−
1
x
)
)
(
12
+
9
x
2
1
+
x
2
)
d
x
, where
tan
−
1
x
takes only principal values, then the value of
(
log
e
|
1
+
α
|
−
3
π
4
)
is
Q.
If
α
=
1
∫
0
(
e
9
x
+
3
tan
−
1
x
)
(
12
+
9
x
2
1
+
x
2
)
d
x
, where
tan
−
1
x
takes only principle values, then the value of
(
ln
|
1
+
α
|
−
3
π
4
)
is
Q.
If
α
=
1
∫
0
(
e
9
x
+
3
tan
−
1
x
)
(
12
+
9
x
2
1
+
x
2
)
d
x
,
Where
tan
−
1
x
takes only principle values, then the value of
(
ln
|
1
+
α
|
−
3
π
4
)
is
Q.
Inverse circular functions,Principal values of
s
i
n
−
1
x
,
c
o
s
−
1
x
,
t
a
n
−
1
x
.
t
a
n
−
1
x
+
t
a
n
−
1
y
=
t
a
n
−
1
x
+
y
1
−
x
y
,
x
y
<
1
π
+
t
a
n
−
1
x
+
y
1
−
x
y
,
x
y
>
1
.
Using the principal values, express the following as a single angle :
3
t
a
n
−
1
(
1
2
)
+
2
t
a
n
−
1
(
1
5
)
+
s
i
n
−
1
142
65
√
5
.
Q.
Inverse circular functions,Principal values of
s
i
n
−
1
x
,
c
o
s
−
1
x
,
t
a
n
−
1
x
.
t
a
n
−
1
x
+
t
a
n
−
1
y
=
t
a
n
−
1
x
+
y
1
−
x
y
,
x
y
<
1
π
+
t
a
n
−
1
x
+
y
1
−
x
y
,
x
y
>
1
.
(a) If
t
a
n
−
1
√
1
+
x
2
−
√
1
−
x
2
√
(
1
+
x
2
)
+
√
(
1
+
x
2
)
=
α
. then prove that
x
2
=
s
i
n
2
α
(b) If
m
t
a
n
(
α
−
θ
)
c
o
s
2
θ
=
n
t
a
n
θ
c
o
s
2
(
α
−
θ
)
, then prove that
θ
=
1
2
[
α
−
t
a
n
−
1
(
n
−
m
n
+
m
t
a
n
α
)
]
(c)
c
o
s
−
1
c
o
s
α
+
c
o
s
β
1
+
c
o
s
α
c
o
s
β
=
2
t
a
n
−
1
(
t
a
n
α
2
t
a
n
β
2
)
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