Given tan (a+b) = 1and.tan (a_b) =1/7.find.tan a and. Tan b
Answers
Answer:
tan
a
=
−
2
,
1
2
;
&
tan
b
=
−
3
,
1
3
.
Explanation:
METHOD I
Let
tan
a
=
x
,
tan
b
=
y
.
Given that,
tan
(
a
+
b
)
=
1
⇒
tan
a
+
tan
b
1
−
tan
a
tan
b
=
1
⇒
x
+
y
=
1
−
x
y
...
(
1
)
Similarly,
tan
(
a
−
b
)
=
1
7
⇒
x
−
y
=
1
7
(
1
+
x
y
)
...
...
...
.
(
2
)
(
1
)
+
(
2
)
⇒
2
x
=
8
7
−
6
7
x
y
,
or,
x
=
4
7
−
3
7
x
y
,
i.e.,
x
(
1
+
3
7
y
)
=
4
7
,
giving,
x
=
4
7
+
3
y
.
We submit this
x
in
(
1
)
to see that.
4
7
+
3
y
+
y
+
(
4
7
+
3
y
)
y
=
1
,
or,
4
+
7
y
+
3
y
2
+
4
y
=
7
+
3
y
,
i.e.,
3
y
2
+
8
y
−
3
=
0
.
Hence,
y
=
tan
b
=
−
3
,
1
3
.
Using
x
=
4
7
+
3
y
,
we get,
x
=
tan
a
=
−
2
,
1
2
.
tan
a
=
−
2
,
1
2
;
tan
b
=
−
3
,
1
3
.
Explanation:
METHOD II:-
Take,
a
+
b
=
C
,
a
−
b
=
D
,
then, by what is given,
tan
C
=
1
,
tan
D
=
1
7
...
(
1
)
Observe that,
C
+
D
=
2
a
.
C
+
D
=
2
a
.
⇒
tan
(
C
+
D
)
=
tan
2
a
.
⇒
tan
C
+
tan
D
1
−
tan
C
tan
D
=
tan
2
a
Here, we use
(
1
)
to get,
1
+
1
7
1
−
1
7
=
tan
2
a
,
or,
tan
2
a
=
4
3
.
.
(
2
)
Recall that
tan
2
a
=
2
tan
a
1
−
tan
2
a
...
...
.
.
(
3
)
.
so, if,
tan
a
=
x
,
(
2
)
&
(
3
)
⇒
2
x
1
−
x
2
=
4
3
,
⇒
3
x
=
2
−
2
x
2
,
⇒
2
x
2
+
3
x
−
2
=
0
,
⇒
(
x
+
2
)
(
2
x
−
1
)
=
0
,
⇒
x
=
tan
a
=
−
2
,
1
2
,
.
as in METHOD I!
tan
b
can similarly be obtained using
C
−
D
=
2
b
.
Answer:
tan a = -2, 1/2
tan b = -3, 1/3
