Question

If $\cos A=\frac{1}{2}$ and $\sin B=\frac{1}{\sqrt{2}}$ , find the value of $\frac{\tan A-\tan B}{1+\tan A\tan B}$

• A. $3-\sqrt{3}$
• B. $2-\sqrt{3}$
• C. $2+\sqrt{3}$
• D. $2-\sqrt{2}$

Answer 1

The correct option is B $2-\sqrt{3}$

$\cos A=\frac{1}{2}$
$\cos A=\cos {60}^0$
$A={60}^{\circ }$
$\sin B=\frac{1}{\sqrt{2}}$
$\sin B=\sin {45}^0$
$B={45}^{\circ }$
Now, $\frac{\tan A-\tan B}{1+\tan A\tan B}$
= $\frac{\tan {60}^0-\tan {45}^0}{1+\tan {60}^0\tan {45}^0}$
= $\frac{\sqrt{3}-1}{1+\left(\sqrt{3}\right)\left(1\right)}$
= $\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$
= $\frac{3+1-2\sqrt{3}}{3-1}$
= $2-\sqrt{3}$