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Let f(x)=tan⁻¹√((1+sin x)/(1-sin x)), where x lies in (0,π/2). The normal to the curve y=f(x) at x=π/6 passes through which point?

1. (π/6, 0)
2. (π/4, 0)
3. (0, 0)
4. (0, 2π/3)

Correct answer: (0, 2π/3)

Solution

For x in (0,pi/2), f(x)=arctan(tan(pi/4 + x/2))=pi/4 + x/2, so f'(x)=1/2. At x=pi/6, f=pi/3 and the normal slope is -2: y-pi/3=-2(x-pi/6). Setting x=0 gives y=pi/3+pi/3=2pi/3, so it passes through (0, 2pi/3).

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