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Question

If 2tan1x=sin12x1+x2, then:

A
xR
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B
x1
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C
x1
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D
1x1
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Solution

The correct option is D 1x1
Putting θ=tan1x,π2<θ<π2, we have
R.H.S. sin12tanθ1+tan2θ=sin1sin2θ=2θ=2tan1x
When π22θπ2 i.e., π4θπ4
i.e., 1x=tanθ1.

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