Question
If 2tan−1x=sin−12x1+x2, then:
Answer: Option D
:
D
Putting θ=tan−1x,−π2<θ<π2, we have
R.H.S. sin−12tanθ1+tan2θ=sin−1sin2θ=2θ=2tan−1x
When −π2≤2θ≤π2 i.e., −π4≤θ≤π4
i.e., −1≤x=tanθ≤1.
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:
D
Putting θ=tan−1x,−π2<θ<π2, we have
R.H.S. sin−12tanθ1+tan2θ=sin−1sin2θ=2θ=2tan−1x
When −π2≤2θ≤π2 i.e., −π4≤θ≤π4
i.e., −1≤x=tanθ≤1.
Was this answer helpful ?
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