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Byju's Answer
Standard XII
Mathematics
General Solution of Trigonometric Equation
θ + tanθ = x ...
Question
cot
θ
+
tan
θ
=
x
and
sec
θ
−
cos
θ
=
y
then
(
x
2
y
)
2
/
3
−
(
x
y
2
)
2
/
3
.
Open in App
Solution
m
=
c
o
t
θ
+
t
a
n
θ
=
c
o
s
θ
s
i
n
θ
+
s
i
n
θ
c
o
s
θ
=
c
o
s
2
θ
+
s
i
n
2
θ
s
i
n
θ
c
o
s
θ
=
c
o
s
e
c
θ
s
e
c
θ
=
m
n
=
s
e
c
θ
−
c
o
s
θ
=
1
c
o
s
θ
−
c
o
s
θ
1
=
1
−
c
o
s
2
θ
c
o
s
θ
=
s
i
n
2
θ
c
o
s
θ
=
t
a
n
θ
s
i
n
Θ
=
n
L
H
S
=
(
m
2
n
)
2
/
3
−
(
m
n
2
)
2
/
3
=
(
m
4
n
2
)
1
/
3
−
(
m
2
n
4
)
1
/
3
=
(
c
o
s
e
c
4
θ
s
e
c
4
θ
t
a
n
2
θ
s
i
n
2
θ
)
1
/
3
−
(
c
o
s
e
c
2
θ
s
e
c
2
θ
t
a
n
4
θ
s
i
n
4
θ
)
1
/
3
=
(
1
s
i
n
4
θ
1
c
o
s
4
θ
s
i
n
2
θ
c
o
s
2
θ
s
i
n
2
θ
)
1
/
3
-
(
1
s
i
n
2
θ
1
c
o
s
2
θ
s
i
n
4
θ
c
o
s
4
θ
s
i
n
4
θ
1
)
1
/
3
=
(
1
c
o
s
6
θ
)
1
/
3
−
(
s
i
n
6
θ
c
o
s
5
6
θ
)
1
/
3
=
s
e
c
2
θ
−
t
a
n
2
θ
=
1
Suggest Corrections
1
Similar questions
Q.
If
cot
θ
+
tan
θ
=
x
and
sec
θ
−
cos
θ
=
y
, then prove that
(
x
2
y
)
2
/
3
−
(
x
y
2
)
2
/
3
=
1
Q.
If
cot
θ
+
tan
θ
=
x
and
sec
θ
−
cos
θ
=
y
, then
Q.
If
c
o
t
θ
+
tan
θ
=
x
and
sec
θ
−
cos
θ
=
y
, then
Q.
If
(
cot
θ
+
tan
θ
)
=
m
and
(
sec
θ
−
cos
θ
)
=
n
then
(
m
2
n
)
2
/
3
−
(
m
n
2
)
2
/
3
=
1
.
Q.
If
cot
θ
+
tan
θ
=
x
and
sec
θ
−
cos
θ
=
y
, then show that
sin
θ
⋅
cos
θ
=
1
x
or
sin
θ
⋅
tan
θ
=
y
.
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