Question

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

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

Answer 1

Answer: (D)

$tan\left(A-B\right)=\frac{tanA-tanB}{1+tanAtanB}$
=$\frac{\frac{\sqrt{3}x}{2K-x}-\frac{2x-K}{\sqrt{3}K}}{1+\frac{\sqrt{3}x}{2K-x}\frac{2x-K}{\sqrt{3}K}}$
$=\frac{3Kx-\left(2x-K\right)\left(2K-x\right)}{\left(2K-x\right)\sqrt{3}K+\sqrt{3x}\left(2x-K\right)}$
$=\frac{3Kx-\left(4Kx-2x^2-2K^2+Kx\right)}{2\sqrt{3}{K}^2-\sqrt{3}Kx+2\sqrt{3}{x}^2-\sqrt{3}Kx}$
$=\frac{2x^2-2Kx+2K^2}{2\sqrt{3}{x}^2-2\sqrt{3}Kx+2\sqrt{3}{k}^2}=\frac{1}{\sqrt{3}}=tan30^{º}$
$\therefore A-B=30^{º}$