If log(a+b+c) = log a + log b + log c then prove that log(2a/1-a^2 + 2b/1-b^2 + 2c/1-c^2) = log 2a/1-a^2 + log 2b/1-b^2 + log 2c/1-c^2
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Answer:
Step-by-step explanation:
Use loga+logb+logc=log(abc)
log
a
+
log
b
+
log
c
=
log
(
a
b
c
)
and then put a=tanA
a
=
tan
A
etc. to find
∑tanA=∏tanA⟹A+B+C=nπ
∑
tan
A
=
∏
tan
A
⟹
A
+
B
+
C
=
n
π
where n
n
is any integer (See here)
and 2a1−a2=2tanA1−tan2A=tan2A
2
a
1
−
a
2
=
2
tan
A
1
−
tan
2
A
=
tan
2
A
⟹∑tan2A=∏tan2A
⟹
∑
tan
2
A
=
∏
tan
2
A
as 2A+
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