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)

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