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Byju's Answer
Standard VII
Mathematics
Properties of Isosceles and Equilateral Triangles
If in Δ ABC...
Question
If in
Δ
A
B
C
,
∠
A
=
sin
−
1
(
x
)
,
∠
B
=
sin
−
1
(
y
)
a
n
d
∠
C
=
sin
−
1
(
z
)
, then
x
√
1
−
y
2
√
1
−
z
2
+
y
√
1
−
x
2
√
1
−
z
2
+
z
√
1
−
x
2
√
1
−
y
2
is :
A
equal to xyz
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B
>
1
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C
equal to
x
2
y
z
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D
none of these
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Solution
The correct option is
A
equal to xyz
The above equation reduces to
s
i
n
A
c
o
s
B
c
o
s
C
+
s
i
n
B
c
o
s
A
c
o
s
C
+
s
i
n
C
c
o
s
A
c
o
s
B
=
c
o
s
C
(
s
i
n
A
c
o
s
B
+
c
o
s
A
s
i
n
B
)
+
s
i
n
C
c
o
s
A
c
o
s
B
=
c
o
s
C
[
s
i
n
(
A
+
B
)
]
+
s
i
n
C
c
o
s
A
c
o
s
B
=
c
o
s
C
[
s
i
n
(
π
−
C
)
]
+
s
i
n
C
c
o
s
A
c
o
s
B
=
c
o
s
C
s
i
n
C
+
s
i
n
C
c
o
s
A
c
o
s
B
=
s
i
n
C
[
c
o
s
C
+
c
o
s
A
c
o
s
B
]
=
s
i
n
C
[
c
o
s
C
+
1
2
(
c
o
s
(
A
+
B
)
+
c
o
s
(
A
−
B
)
)
]
=
s
i
n
C
[
c
o
s
C
+
1
2
(
c
o
s
(
π
−
C
)
+
c
o
s
(
A
−
B
)
)
]
=
s
i
n
C
[
c
o
s
C
+
−
1
2
c
o
s
(
C
)
+
1
2
c
o
s
(
A
−
B
)
]
=
s
i
n
C
[
1
2
c
o
s
(
C
)
+
1
2
c
o
s
(
A
−
B
)
]
=
s
i
n
C
[
1
2
c
o
s
(
A
−
B
)
−
1
2
c
o
s
(
A
+
B
)
]
=
s
i
n
A
s
i
n
B
s
i
n
C
=
x
y
z
Suggest Corrections
0
Similar questions
Q.
Prove the following:
s
i
n
−
1
x
+
s
i
n
−
1
y
+
s
i
n
−
1
z
=
π
Show that
x
√
1
−
x
2
+
y
√
1
−
y
2
+
z
√
1
−
z
2
=
x
y
z
Q.
If
sin
−
1
x
+
sin
−
1
y
+
sin
−
1
z
=
π
, prove that
x
√
1
−
x
2
+
y
√
1
−
y
2
+
z
√
1
−
z
2
=
2
x
y
z
.
Q.
Inverse circular functions,Principal values of
s
i
n
−
1
x
,
c
o
s
−
1
x
,
t
a
n
−
1
x
.
t
a
n
−
1
x
+
t
a
n
−
1
y
=
t
a
n
−
1
x
+
y
1
−
x
y
,
x
y
<
1
π
+
t
a
n
−
1
x
+
y
1
−
x
y
,
x
y
>
1
.
(a) If
c
o
s
−
1
p
+
c
o
s
−
1
q
+
c
o
s
−
1
r
=
π
, then prove that
p
2
+
q
2
+
r
2
+
2
p
q
r
=
1
(b) If
s
i
n
−
1
x
+
s
i
n
−
1
y
+
s
i
n
−
1
z
=
π
, then prove that
x
4
+
y
4
+
z
4
+
4
x
2
y
2
z
2
=
2
(
x
2
y
2
+
y
2
z
2
+
z
2
x
2
)
(c) If
t
a
n
−
1
x
+
t
a
n
−
1
y
+
t
a
n
−
1
z
=
π
or
π
/
2
show that
x
+
y
+
z
=
x
y
z
or
x
y
+
y
z
+
z
x
=
1
.
Q.
If
sin
−
1
x
+
sin
−
1
y
+
sin
−
1
z
=
π
, then prove that
x
√
1
−
x
2
+
y
√
1
−
y
2
+
z
√
1
−
z
2
=
2
x
y
z
Q.
If
sin
−
1
x
+
sin
−
1
y
+
sin
−
1
z
=
π
2
,
then the value of
x
2
+
y
2
+
z
2
+
2
x
y
z
is equal to
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