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Tardigrade
Question
Mathematics
If sec θ=m and tan θ=n, then (1/m)[(m+n)+(1/(m+n))]=
Q. If
sec
θ
=
m
and
tan
θ
=
n
, then
m
1
[
(
m
+
n
)
+
(
m
+
n
)
1
]
=
1715
149
Report Error
A
2
B
-1
C
1
D
0
Solution:
sec
θ
=
m
,
tan
θ
=
n
m
1
[
(
m
+
n
)
+
m
+
n
1
]
=
s
e
c
θ
1
[
sec
θ
+
tan
θ
+
s
e
c
θ
+
t
a
n
θ
1
]
=
s
e
c
θ
1
[
sec
θ
+
tan
θ
+
sec
θ
−
tan
θ
]
=
s
e
c
θ
2
s
e
c
θ
=
2