Question

If $tan^{-1}\left(\frac{x\sqrt{3}}{2K-x}\right) and B=tan^{-1}\left(\frac{2x-K}{K\sqrt{3}}\right)$, then the value of $A - B$ is

• A. $0^º$
• B. $45^º$
• C. $60^º$
• D. $30^º$

Answer 1

Answer: (D)

$tan\left(A-B\right)=\frac{tan\,A - tan\, B}{1+tan\,A\,tan\,B}$
=$\frac{\frac{\sqrt{3}x}{2K-x}-\frac{2x-K}{\sqrt{3}K}}{1+\frac{\sqrt{3}x}{2K-x} \frac{2x-K}{\sqrt{3}K}}$
$=\frac{3Kx-\left(2x-K\right)\left(2K-x\right)}{\left(2K-x\right)\sqrt{3}K+\sqrt{3x}\left(2x-K\right)}$
$=\frac{3Kx-\left(4Kx-2x^{2}-2K^{2}+Kx\right)}{2\sqrt{3}K^{2}-\sqrt{3}Kx+2\sqrt{3}x^{2}- \sqrt{3}Kx}$
$=\frac{2x^{2}-2Kx+2K^{2}}{2\sqrt{3}x^{2}-2\sqrt{3}Kx+2\sqrt{3}k^{2}}=\frac{1}{\sqrt{3}}=tan\,30^{º}$
$\therefore A-B=30^{º}$