Answer
$$\tan67.5^\circ=\sqrt2+1$$
9 should be matched with F.
Work Step by Step
$$\tan67.5^\circ$$
$67.5^\circ$ is the half angle of $135^\circ$
$$67.5^\circ=\frac{135^\circ}{2}$$
Therefore, $$\tan67.5^\circ=\tan\Big(\frac{135^\circ}{2}\Big)$$
- Recall the half-angle identities:
$$\tan\Big(\frac{A}{2}\Big)=\frac{\sin A}{1+\cos A}$$
Apply the identity to $\tan67.5^\circ$ with $A=135^\circ$, we have
$$\tan67.5^\circ=\frac{\sin135^\circ}{1+\cos135^\circ}$$
As $135^\circ+45^\circ=180^\circ$, $|\sin135^\circ|=\sin45^\circ$ and $|\cos135^\circ|=\cos45^\circ$
$135^\circ$ lies in quadrant II, where $\sin\theta\gt0$ and $\cos\theta\lt0$.
Therefore, $\sin135^\circ=\sin45^\circ=\frac{\sqrt2}{2}$ and $\cos135^\circ=-\cos45^\circ=-\frac{\sqrt2}{2}$
So, $$\tan67.5^\circ=\frac{\frac{\sqrt2}{2}}{1-\frac{\sqrt2}{2}}$$
$$\tan67.5^\circ=\frac{\frac{\sqrt2}{2}}{\frac{2-\sqrt2}{2}}$$
$$\tan67.5^\circ=\frac{\sqrt2}{2-\sqrt2}$$
$$\tan67.5^\circ=\frac{\sqrt2}{\sqrt2(\sqrt2-1)}$$
$$\tan67.5^\circ=\frac{1}{\sqrt2-1}$$
Multiply both numerator and denominator with $\sqrt2+1$.
- Numerator: $1\times(\sqrt2+1)=\sqrt2+1$
- Denominator: $(\sqrt2-1)(\sqrt2+1)=2-1=1$
Thus, $$\tan67.5^\circ=\frac{\sqrt2+1}{1}$$
$$\tan67.5^\circ=\sqrt2+1$$
9 should be matched with F.
