128
If $\sec A = \sqrt{2}$ and $\tan B = \sqrt{3}$, then find the value of $2 \sin A \cos B$.
Show SolutionHide Solution↓
$\sec A = \sqrt{2} \Rightarrow A = 45^{\circ}$, $\tan B = \sqrt{3} \Rightarrow B = 60^{\circ}$ (1/2 Mark + 1/2 Mark)
$2 \sin A \cos B = 2 \sin 45^{\circ} \cos 60^{\circ} = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$ or $\frac{\sqrt{2}}{2}$ (1 Mark)
$2 \sin A \cos B = 2 \sin 45^{\circ} \cos 60^{\circ} = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$ or $\frac{\sqrt{2}}{2}$ (1 Mark)