If tan (theta/2) = tan (fi/2) and tan fi = 2 tan alpha
then
prove that theta + fi = 2 alpha
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Answer:
Let us put tanθ=t
1
,tanϕ=t
2
∴t
1
2
t
2
2
=(
a+b
a−b
)
or (a+b)t
1
2
t
2
2
=a−b ....(1)
Also cos2θ=
1+tan
2
θ
1−tan
2
θ
=
1+t
1
2
1−t
1
2
etc.
Now a - b cos 2θ=a−b
(1+t
1
2
)
(1−t
1
2
)
=
(1+t
1
2
)
(a−b)+(a+b)t
1
2
Put for (a - b) from (1),
=
(1+t
1
2
)
a+b
[t
1
2
t
2
2
+t
1
2
]=
(1+t
1
2
)
(a+b)t
1
2
(1+t
2
2
)
,
Similarly, a - b cos 2ϕ=
1+t
2
2
(a+b)
t
2
2
(1+t
1
2
)
∴ (a - b cos 2θ) (a - b cos 2ϕ) = (a+b)
2
t
1
2
t
2
2
=(a+b)
2
{(a−b)/(a+b)}=a
2
−b
2
,
which is independent of θ.
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