Question

If $\tan A=1$ and $\tan B=\sqrt{3}$ evaluate:$\cos A\cos B-\sin A\sin B$

• A. $\frac{\sqrt{2}+\sqrt{6}}{4}$
• B. $\frac{\sqrt{2}-\sqrt{6}}{6}$
• C. $\frac{1-\sqrt{3}}{2\sqrt{2}}$
• D. $\frac{\sqrt{2}+\sqrt{6}}{6}$

Answer 1

Answer: (C)

$\frac{1-\sqrt{3}}{2\sqrt{2}}$

$\tan A=1$ and $\tan B=\sqrt{3}$

$\tan A=\tan 45$ and $\tan B=\tan 60$

$A={45}^{\circ }$ and $B={60}^{\circ }$

Now,

= $\cos A\cos B-\sin A\sin B$

= $\cos {45}^0\cos {60}^0-\sin {45}^0\sin {60}^0$

= $\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)-\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)$

= $\frac{1}{2\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}}$

= $\frac{1}{2\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}}$

= $\frac{1-\sqrt{3}}{2\sqrt{2}}$