cot−1√1+sin + √1−sin√1+sin − √1−sin = 2 , ∈ (0,4)
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Answer:
LHS =cot
−1
[
1+sinx
−
1−sinx
1+sinx
+
1−sinx
],x∈(0,
4
π
)
Given, 0<x<
4
π
⇒0<x<
8
π
⇒
2
x
∈(0,
4
π
)⊂(0,π)
cot
−1
⎝
⎜
⎜
⎜
⎛
(cos
2
x
+sin
2
x
)
2
+
(cos
2
x
−sin
2
x
)
2
(cos
2
x
+sin
2
x
)
2
+
(cos
2
x
−sin
2
x
)
2
⎠
⎟
⎟
⎟
⎞
=cot
−1
⎝
⎛
cos
2
x
+sin
2
x
−cos
2
x
+sin
2
x
cos
2
x
+sin
2
x
+cos
2
x
−sin
2
x
⎠
⎞
=cot
−1
⎝
⎛
sin
2
x
cos
2
x
⎠
⎞
⇒cot
−1
(cot
2
x
)⇒
2
x
= RHS
Hence proved
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