Question

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

• A. $\frac{1}{\sqrt{3}}$
• B. $\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{3}$

Answer 1

The correct option is D $\sqrt{3}$

Given : $\frac{\cos (15°)+\sin (15°)}{\cos (15°)-\sin (15°)}$
Divide the given expression by in numerator and dinominator by $\cos (15°)$
$\Rightarrow \frac{1+\tan (15°)}{1-\tan (15°)}=\frac{1+\tan (15°)}{1-\left(1\right)\left(\tan \right(15°\left)\right)}\Rightarrow \frac{\tan (45°)+\tan (15°)}{1-\tan (45°)\tan (15°)}\Rightarrow \tan (45°+15°)\Rightarrow \tan (60°)=\sqrt{3}$
alternate ,
we know that,
$\cos (15°)=\frac{\sqrt{3}+1}{2\sqrt{2}}\&\sin (15°)=\frac{\sqrt{3}-1}{2\sqrt{2}}$we have,
$\frac{\cos (15°)+\sin (15°)}{\cos (15°)-\sin (15°)}\Rightarrow \frac{\frac{\sqrt{3}+1}{2\sqrt{2}}+\frac{\sqrt{3}-1}{2\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{\sqrt{3}-1}{2\sqrt{2}}}\Rightarrow \sqrt{3}$