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AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper
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© examsnet.com
Question : 21
Total: 160
If ABC is not a right angled triangle and sin
(
π
4
−
A
)
sin
(
π
4
−
B
)
= -
1
2
√
2
cosec
(
π
4
−
C
)
, then tan A tan B + tan B tan C + tan C tan A =
cot A + cot B + cot C
tan A + tan B + tan C
1
t
a
n
A
+
t
a
n
B
+
t
a
n
C
1
c
o
t
A
+
c
o
t
B
+
c
o
t
C
Validate
Solution:
Consider the expression,
A
+
B
+
C
=
π
Then,
sin
(
π
4
−
A
)
sin
(
π
4
−
B
)
sin
(
π
4
−
C
)
=
−
1
2
√
2
(
1
√
3
)
3
(
cos
A
−
sin
A
)
(
cos
B
−
sin
B
)
(
cos
C
−
sin
C
)
=
−
1
2
√
2
cos
A
cos
B
cos
C
(
1
−
tan
A
)
(
1
−
tan
B
)
(
1
−
tan
C
)
=
−
1
cos
A
cos
B
cos
C
(
1
−
tan
A
−
tan
B
−
tan
C
+
tan
A
tan
B
+
tan
B
tan
C
+
tan
A
tan
C
−
tan
A
tan
B
tan
C
)
=
cos
A
+
B
+
C
[
∵
A
+
B
+
C
=
π
]
Further simplify above,
−
2
(
tan
A
+
tan
B
+
tan
C
)
=
−
2
(
tan
A
tan
B
+
tan
B
tan
C
+
tan
A
tan
C
)
tan
A
tan
B
+
tan
B
tan
c
+
tan
A
tan
C
=
tan
A
+
tan
B
+
tan
C
© examsnet.com
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