Question

Calculate without resorting to table.
$\frac{\cot {15}^{\circ }+1}{2\cot {15}^{\circ }}$

Answer 1

$\frac{\cot 15°+1}{2\cot 15°}$

$=\frac{\sin 15°+\cos 15°}{2\cot 15°}$

$=\frac{1+\tan 15°}{2}$

We know that,

$\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$

$\tan 30°=\frac{2\tan 15°}{1-{\tan }^{2}15°}$

Let $\tan 15°=z$

$\frac{1}{\sqrt{3}}=\frac{2z}{1-z^2}$

$1-z^2=2\sqrt{3}z$

$z^2-2\sqrt{3}z-1=0$

${(z+\sqrt{3})}^2=4$

${(z+\sqrt{3})}^2=4$

$z=2-\sqrt{3}=\tan 15°$

Now, $\frac{1+\tan 15°}{2}=\frac{3+\sqrt{3}}{2}=\frac{\sqrt{3}(\sqrt{3}+1)}{2}$