Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The number of principal solutions of tan 2 θ = 1 is
Q. The number of principal solutions of
tan
2
θ
=
1
is
3092
118
MHT CET
MHT CET 2017
Trigonometric Functions
Report Error
A
One
14%
B
Two
57%
C
Three
10%
D
Four
19%
Solution:
Let
y
=
tan
2
θ
tan
2
θ
=
1
⟹
2
θ
=
tan
−
1
(
1
)
⟹
2
θ
=
4
π
+
kπ
, where
k
is a positive integer
Now,
for
k
=
0
⇒
θ
=
4
π
and for
k
=
1
⇒
θ
=
4
5
π
0
≤
4
π
≤
2
π
and
0
≤
4
5
π
≤
2
π
∴
2
θ
=
4
π
,
4
5
π
⟹
θ
=
8
π
,
8
5
π
Hence, the required principal solutions are
8
π
and
8
5
π