Question

If $\tan A=1$ and $\tan B=\sqrt{3},$ then evaluate: $\sin A\cos B+\cos A\sin B$.

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

Answer 1

The correct option is D $\frac{\sqrt{2}+\sqrt{6}}{4}$

$\tan A=1$
$\tan A=\tan 45$
$A={45}^{\circ }$
$\tan B=\sqrt{3}$
$\tan B=\tan 60$
$B={60}^{\circ }$
Now, $\sin A\cos B+\cos A\sin B$

= $\sin \left({45}^0\right)\cos \left({60}^0\right)+cos\left({45}^0\right)\sin \left({60}^0\right)$

= $\frac{1}{\sqrt{2}}.\frac{1}{2}+\frac{1}{\sqrt{2}}.\frac{\sqrt{3}}{2}$

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

= $\frac{1+\sqrt{3}}{2\sqrt{2}}$
Multiply and divide by $\sqrt{2}$

= $\frac{\sqrt{2}+\sqrt{6}}{4}$