Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If tan 15°+(1/ tan 75°)+(1/ tan 105°)+ tan 195°=2 a, then the value of (a+(1/a)) is :
Q. If
tan
1
5
∘
+
t
a
n
7
5
∘
1
+
t
a
n
10
5
∘
1
+
tan
19
5
∘
=
2
a
, then the value of
(
a
+
a
1
)
is :
1836
126
JEE Main
JEE Main 2023
Trigonometric Functions
Report Error
A
4
51%
B
2
13%
C
4
−
2
3
24%
D
5
−
2
3
3
12%
Solution:
tan
1
5
∘
=
2
−
3
t
a
n
7
5
∘
1
=
cot
7
5
∘
=
2
−
3
t
a
n
10
5
∘
1
=
cot
(
10
5
∘
)
=
−
cot
7
5
∘
=
3
−
2
tan
19
5
∘
=
tan
1
5
∘
=
2
−
3
∴
2
(
2
−
3
)
=
2
a
⇒
a
=
2
−
3
⇒
a
+
a
1
=
4