Question

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

• A. $0^{\circ} $
• B. $45^{\circ} $
• C. $60^{\circ} $
• D. $30^{\circ} $

Answer 1

Answer: (D)

$A= \tan^{-1} \left(\frac{x\sqrt{3}}{2k -x}\right),B = \tan ^{-1}\left(\frac{2x -k}{k\sqrt{3}}\right)$
$ A-B =\tan ^{-1} \left(\frac{x\sqrt{3}}{2k -x}\right) -\tan ^{-1} \left(\frac{2x-k}{k\sqrt{3}}\right)$
$ = \tan ^{-1} \left(\frac{\frac{x\sqrt{3}}{2k -x} -\frac{2x -k}{k\sqrt{3}}}{1+ \left(\frac{x\sqrt{3}}{2k -x} -\frac{2x -k}{k\sqrt{3}}\right)}\right) $
$=\tan ^{-1} \left[\frac{3xk-\left(2k -x\right)\left(2x-k\right)}{\left(2k-x\right)k\sqrt{3} +x\sqrt{3} \left(2x -k\right)}\right] =$
$ \tan ^{-1} \left(\frac{-2 kx + 2k^{2} + 2x^{2}}{2\sqrt{3} \left(k^{2} +x^{2}\right) -2 \sqrt{3} xk}\right) $
$= \tan ^{-1} \left[\frac{2\left(k^{2} +x^{2} -kx\right)}{2\sqrt{3} \left(k^{2} +x^{2} -kx\right)}\right] $
$= \tan ^{-1} \left(\frac{1}{\sqrt{3}}\right) = 30^{\circ}$