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Byju's Answer
Standard XII
Mathematics
Solving Simultaneous Trigonometric Equations
Prove that : ...
Question
Prove that :
sin
−
1
x
+
sin
−
1
y
=
sin
−
1
[
x
√
1
−
y
2
+
y
√
1
−
x
2
]
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Solution
Suppose that
sin
−
1
x
=
α
⇒
x
=
sin
α
and
sin
−
1
y
=
β
⇒
y
=
sin
β
Now
sin
(
α
+
β
)
=
sin
α
.
cos
β
+
cos
α
.
sin
β
=
sin
(
α
+
β
)
=
sin
α
.
√
1
−
sin
2
β
+
√
1
−
sin
2
α
.
sin
β
=
sin
(
α
+
β
)
=
x
.
√
1
−
y
2
+
y
.
√
1
−
x
2
=
α
+
β
=
sin
−
1
[
x
.
√
1
−
y
2
+
y
.
√
1
−
x
2
]
=
sin
−
1
x
+
sin
−
1
y
=
sin
−
1
[
x
√
1
−
y
2
+
y
√
1
−
x
2
]
.
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0
Similar questions
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.
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
=
tan
−
1
y
then prove that
1
x
2
−
1
y
2
=
1
.
Q.
Inverse circular functions,Principal values of
sin
−
1
x
,
cos
−
1
x
,
tan
−
1
x
.
tan
−
1
x
+
tan
−
1
y
=
tan
−
1
x
+
y
1
−
x
y
,
x
y
<
1
π
+
tan
−
1
x
+
y
1
−
x
y
,
x
y
>
1
.
Prove that
tan
−
1
1
−
x
1
+
x
tan
−
1
1
−
y
1
+
y
=
sin
−
1
y
−
x
√
1
+
x
2
√
1
+
y
2
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
.
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