If sinA+sinB=x and cosA+CosB=y prove that tan(A+B/2)=x/y and cos (A-B/2)=√xpower2+ypower2/2
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Step-by-step explanation:
Explanation:
sin
A
+
sin
B
=
x
⇒
2
⋅
sin
(
A
+
B
2
)
⋅
cos
(
A
−
B
2
)
=
x
...
(
1
)
cos
A
+
cos
B
=
y
⇒
2
⋅
cos
(
A
+
B
2
)
⋅
cos
(
A
−
B
2
)
=
y
...
(
2
)
Now, we have to remove the term of sine and cosine value of
(
A
+
B
2
)
to get the value of
tan
(
A
−
B
2
)
.
From 1st equation,
⇒
4
⋅
sin
2
(
A
+
B
2
)
⋅
cos
2
(
A
−
B
2
)
=
x
2
⇒
4
⋅
cos
2
(
A
−
B
2
)
⋅
{
1
−
cos
2
(
A
+
B
2
)
}
=
x
2
...
(
3
)
Similarly, from 2nd equation,
⇒
4
⋅
cos
2
(
A
+
B
2
)
⋅
cos
2
(
A
−
B
2
)
=
y
2
⇒
4
⋅
cos
2
(
A
+
B
2
)
⋅
cos
2
(
A
−
B
2
)
=
y
2
...
(
4
)
From 3rd and 4th equation, we get
⇒
4
⋅
cos
2
(
A
−
B
2
)
⋅
⎧
⎪
⎨
⎪
⎩
1
−
y
2
4
⋅
cos
2
(
A
−
B
2
)
⎫
⎪
⎬
⎪
⎭
=
x
2
⇒
4
⋅
cos
2
(
A
−
B
2
)
−
y
2
=
x
2
⇒
4
⋅
cos
2
(
A
−
B
2
)
=
x
2
+
y
2
⇒
cos
2
(
A
−
B
2
)
=
x
2
+
y
2
4
⇒
sec
2
(
A
−
B
2
)
=
4
x
2
+
y
2
⇒
1
+
tan
2
(
A
−
B
2
)
=
4
x
2
+
y
2
⇒
tan
2
(
A
−
B
2
)
=
4
x
2
+
y
2
−
1
⇒
tan
2
(
A
−
B
2
)
=
4
−
x
2
−
y
2
x
2
+
y
2
⇒
tan
(
A
−
B
2
)
=
±
√
4
−
x
2
−
y
2
x
2
+
y
2
Hope it helps..
Thank you...
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