If A=Tan-1x32k-x and B=Tan-12x-kk3 then A–B =

tan A=x32k-x tan B=2x-kk3tan (A-B)=x32k-x-2x-kk31+x32k-x.2x-kk3 =3kx-4kx+2k2+2x2-kx2k23-kx3+2x23-kx3 =2(k2+x2-2kx)23(k2+x2-2kx)=13A-B=π6

If A=Tan-1x32k-x and B=Tan-12x-kk3 then A–B =

tan A=x32k-x tan B=2x-kk3tan (A-B)=x32k-x-2x-kk31+x32k-x.2x-kk3 =3kx-4kx+2k2+2x2-kx2k23-kx3+2x23-kx3 =2(k2+x2-2kx)23(k2+x2-2kx)=13A-B=π6

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If $A = {Tan}^{- 1} (\frac{x \sqrt{3}}{2 k - x})$ and $B = {Tan}^{- 1} (\frac{2 x - k}{k \sqrt{3}})$ then A–B =

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$\frac{π}{2}$

$\frac{π}{6}$

$\frac{π}{3}$

$\frac{π}{4}$

answer is C.

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Detailed Solution

$\begin{array}{rcl} \tan A & = & \frac{x \sqrt{3}}{2 k - x} \tan B = \frac{2 x - k}{k \sqrt{3}} \\ \tan (A - B) & = & \frac{\frac{x \sqrt{3}}{2 k - x} - \frac{2 x - k}{k \sqrt{3}}}{1 + \frac{x \sqrt{3}}{2 k - x} . \frac{2 x - k}{k \sqrt{3}}} \\ = & \frac{3 kx - 4 kx + 2 k^{2} + 2 x^{2} - kx}{2 k^{2} \sqrt{3} - kx \sqrt{3} + 2 x^{2} \sqrt{3} - kx \sqrt{3}} \\ = & \frac{2 (k^{2} + x^{2} - 2 kx)}{2 \sqrt{3} (k^{2} + x^{2} - 2 kx)} = \frac{1}{\sqrt{3}} \\ A - B & = & \frac{π}{6} \end{array}$