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Tardigrade
Question
Mathematics
tan 9°- tan 27°- tan 63°+ tan 81°=
Q.
tan
9
∘
−
tan
2
7
∘
−
tan
6
3
∘
+
tan
8
1
∘
=
1768
194
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A
1
B
2
C
3
D
4
Solution:
tan
9
∘
−
tan
2
7
∘
−
tan
6
3
∘
+
tan
8
1
∘
=
tan
8
1
∘
+
tan
9
∘
−
(
tan
6
3
∘
+
tan
2
7
∘
)
=
cot
9
∘
+
tan
9
∘
−
(
cot
2
7
∘
+
tan
2
7
∘
)
=
s
i
n
9
∘
⋅
c
o
s
9
∘
c
o
s
2
9
∘
+
s
i
n
2
9
∘
−
s
i
n
27
⋅
c
o
s
27
c
o
s
2
2
7
∘
+
s
i
n
2
2
7
∘
=
s
i
n
9
∘
⋅
c
o
s
9
∘
1
−
s
i
n
2
7
∘
⋅
c
o
s
2
7
∘
1
=
s
i
n
1
8
∘
2
−
s
i
n
5
4
∘
2
=
2
s
i
n
54
⋅
s
i
n
18
s
i
n
54
−
s
i
n
18
=
c
o
s
36
s
i
n
18
4
c
o
s
36
⋅
s
i
n
18
=
4