Sec A + Tan A = 2, find sec A
Answers
Given equation
sec
x
+
tan
x
=
2
...
...
.
.
[
1
]
Again we know
sec
2
x
−
tan
2
x
=
1
...
...
.
[
2
]
Dividing [2] by [1]
sec
x
−
tan
x
=
1
2
...
...
.
[
3
]
Adding [1] and [3] we get
2
sec
x
=
2
+
1
2
=
5
2
⇒
sec
x
=
5
4
⇒
cos
x
=
4
5
=
cos
α
,where
α
=
cos
−
1
(
4
5
)
So
x
=
2
n
π
±
α
where
n
∈
Z
Again subtracting [3] from [1] we get a solution in another form
2
tan
x
=
3
2
tan
x
=
3
4
=
tan
β
,where
β
=
tan
−
1
(
3
4
)
So
x
=
n
π
+
β
where
n
∈
Z
Alternatve
Given equation
sec
x
+
tan
x
=
2
⇒
1
cos
x
+
sin
x
cos
x
=
2
⇒
(
√
1
+
sin
x
)
2
√
(
1
+
sin
x
)
(
1
−
sin
x
)
=
2
[
1
+
sin
x
≠
0
]
⇒
√
1
+
sin
x
√
1
−
sin
x
=
2
⇒
1
+
sin
x
1
−
sin
x
=
4
⇒
(
1
+
sin
x
)
=
4
(
1
−
sin
x
)
⇒
5
sin
x
=
3
⇒
sin
x
=
3
5
=
sin
sin
−
1
(
3
5
)
⇒
x
=
n
π
+
(
−
1
)
n
sin
−
1
(
3
5
)