Question

Let $\tan A=\frac{\sqrt{3}}{4-\sqrt{3}}and\tan B=\frac{\sqrt{3}}{4+\sqrt{3}}$ then
$\frac{\tan A-\tan B}{1+\tan A\tan B}$ is equal to

• A. 0.385
• B. 0.375
• C. 0.575
• D. 0.465

Answer 1

The correct option is B 0.375

Given: $\tan A=\frac{\sqrt{3}}{4-\sqrt{3}}and\tan B=\frac{\sqrt{3}}{4+\sqrt{3}}$
Now, $\frac{\tan A-\tan B}{1+\tan A\tan B}$
= $\frac{\frac{\sqrt{3}}{4-\sqrt{3}}-\frac{\sqrt{3}}{4+\sqrt{3}}}{1+\left(\frac{\sqrt{3}}{4-\sqrt{3}}\right)\left(\frac{\sqrt{3}}{4+\sqrt{3}}\right)}$
= $\frac{4\sqrt{3}+3-4\sqrt{3}+3}{16-3+3}$
= $\frac{6}{16}$
= $0.375$