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Tardigrade
Question
Mathematics
The minimum value of 27 tan 2 θ+3 cot 2 θ is
Q. The minimum value of
27
tan
2
θ
+
3
cot
2
θ
is
2172
225
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EAMCET 2012
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A
15
B
18
C
24
D
30
Solution:
Given trigonometrical equation is
27
tan
2
θ
+
3
cot
2
θ
(Using
A
M
≥
GM
)
∴
2
27
t
a
n
2
θ
+
3
c
o
t
2
θ
≥
27
tan
2
θ
⋅
3
cot
2
θ
⇒
2
27
t
a
n
2
θ
+
3
c
o
t
2
θ
≥
81
⇒
27
tan
2
θ
+
3
cot
2
θ
≥
18
Hence, minimum value of given equation is 18