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Byju's Answer
Standard XII
Mathematics
Graph of Trigonometric Ratios
In Δ ABC if...
Question
In
Δ
A
B
C
if
cot
A
2
.
cot
B
2
=
c
;
cot
B
2
cot
C
2
=
a
;
and
cot
C
2
cot
A
2
=
b
then
1
s
−
a
+
1
s
−
b
+
1
s
−
c
=
A
-1
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B
0
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C
1
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D
2
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Solution
The correct option is
D
2
tan
A
/
2
=
Δ
(
s
−
a
)
s
s
tan
A
/
2
Δ
=
1
(
s
−
a
)
S
Δ
(
tan
A
/
2
+
tan
B
/
2
+
tan
C
/
2
)
cot
A
/
2
=
(
s
−
a
)
s
Δ
,
cot
B
/
2
=
(
s
−
b
)
s
Δ
,
cot
C
/
2
=
(
s
−
c
)
s
Δ
(
s
−
c
)
(
s
−
c
)
×
(
s
−
a
)
s
Δ
×
(
s
−
b
)
s
Δ
=
C
s
(
s
−
c
)
=
C
1
(
s
−
c
)
=
c
s
, ---(i) Similarly
1
(
s
−
b
)
=
b
s
,
.
.
.
(
i
i
)
1
(
s
−
a
)
=
a
s
....(iii)
after observing eq(i),eq(ii) and eq(iii)
So the equation asked in the qs is equal to
c
s
+
b
s
+
a
s
=
a
+
b
+
c
s
=
2
.
Suggest Corrections
0
Similar questions
Q.
In a triangle
A
B
C
,
if
cot
(
A
2
)
cot
(
B
2
)
=
c
,
cot
(
B
2
)
cot
(
C
2
)
=
a
,
and
cot
(
C
2
)
cot
(
A
2
)
=
b
, then
1
s
−
a
+
1
s
−
b
+
1
s
−
c
=
Q.
In a triangle
A
B
C
if
cot
A
2
cot
B
2
=
c
,
cot
B
2
cot
C
2
=
a
and
cot
C
2
cot
A
2
=
b
then
1
s
−
a
+
1
s
−
b
+
1
s
−
c
=
Q.
In
Δ
ABC, if
cot
A
2
⋅
cot
B
2
=
c
,
cot
B
2
⋅
cot
C
2
=
a
and
cot
C
2
⋅
cot
A
2
=
b
then
1
s
−
a
+
1
s
−
b
+
1
s
−
c
=
.
Q.
If a, b, c are sides of a triangle and
∣
∣ ∣ ∣
∣
a
2
b
2
c
2
(
a
+
1
)
2
(
b
+
1
)
2
(
c
+
1
)
2
(
a
−
1
)
2
(
b
−
1
)
2
(
c
−
1
)
2
∣
∣ ∣ ∣
∣
=
0
, then
Q.
Statement 1: In a
Δ
A
B
C
, if
b
+
c
=
3
a
, then
cot
B
2
cot
C
2
=
2
Statement 2: In
a
Δ
A
B
C
, if
b
+
c
a
=
m
n
(
m
>
n
and
m
,
n
are positive)
then
cot
B
2
cot
C
2
=
m
+
n
m
−
n
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