Question

If $\cos A=0.5$ and $\cos B=\frac{1}{\sqrt{2}}$;find the value of $\frac{\tan A-\tan B}{1+\tan A\tan B}$

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

Answer 1

Answer: (D)

$2-\sqrt{3}$

$\cos A=0.5$
$\cos A=\cos 30$
$A={30}^{\circ }$
$\cos B=\frac{1}{\sqrt{2}}$
$\cos B=\cos 45$
$B={45}^{\circ }$
Now, $\frac{\tan 30-\tan 45}{1+\tan 30\tan 45}$
= $\frac{\tan 30-\tan 45}{1+\tan 30\tan 45}$
= $\frac{\frac{1}{\sqrt{3}}-1}{1+\left(\frac{1}{\sqrt{3}}\right)\left(1\right)}$
= $\frac{1-\sqrt{3}}{\sqrt{3}+1}$
= $\frac{(1-\sqrt{3})(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$
= $\frac{3+1-2\sqrt{3}}{3-1}$
= $2-\sqrt{3}$