Question

The value of $\frac{\cos (15°)+\sin (15°)}{\cos (15°)-\sin (15°)}$ is

Answer 1

Given expression as:
$\frac{\cos (15°)+\sin (15°)}{\cos (15°)-\sin (15°)}$
Now, Dividing the given expression by $\cos {15}^{c}irc$ in both numerator and dinominator, we get the expression as:

$\frac{1+\tan (15°)}{1-\tan (15°)}=\frac{1+\tan (15°)}{1-\left(1\right)\left(\tan \right(15°\left)\right)}=\frac{\tan (45°)+\tan (15°)}{1-\tan (45°)\tan (15°)}=\tan (45°+15°)=\tan (60°)=\sqrt{3}$
Method 2: We know that:

$\cos (15°)=\frac{\sqrt{3}+1}{2\sqrt{2}}\&\sin (15°)=\frac{\sqrt{3}-1}{2\sqrt{2}}$$\Rightarrow \frac{\cos (15°)+\sin (15°)}{\cos (15°)-\sin (15°)}=\frac{\frac{\sqrt{3}+1}{2\sqrt{2}}+\frac{\sqrt{3}-1}{2\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{\sqrt{3}-1}{2\sqrt{2}}}=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{3}+1-\sqrt{3}+1}=\sqrt{3}$