Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If |tan A + cot A = 2, then the value of tan4 A + cot 4 A =
Q. If
∣
t
an
A
+
cot
A
=
2
, then the value of
tan
4
A
+
cot
4
A
=
8481
199
KCET
KCET 2020
Trigonometric Functions
Report Error
A
2
37%
B
1
19%
C
4
33%
D
5
11%
Solution:
tan
A
+
cot
A
=
2
⇒
(
tan
A
+
cot
A
)
2
=
4
⇒
tan
2
A
+
cot
2
A
+
2
tan
A
cot
A
=
4
⇒
tan
2
A
+
cot
2
A
=
2
⇒
(
tan
2
A
+
cot
2
A
)
2
=
4
⇒
tan
4
A
+
cot
4
A
+
2
tan
2
A
cot
2
A
=
4
⇒
tan
4
A
+
cot
4
A
=
2