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. $0$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is B $\frac{\pi }{6}$

Given: $A={\tan }^{-1}\left(\frac{x\sqrt{3}}{2k-x}\right)$

Formula : ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right),x,y>0$

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

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

$={\tan }^{-1}\left[\frac{3kx-[4kx-2k^2-2x^2+kx]}{\sqrt{3}(2k^2-kx+2x^2-kx)}\right]$

$={\tan }^{-1}\left[\frac{2(k^2+x^2-kx)}{2\sqrt{3}(k^2+x^2-kx)}\right]$

$\Rightarrow A-B={\tan }^{-1}\frac{1}{\sqrt{3}}=\frac{\pi }{6}$