Question

Find the value of the following:
$\frac{\cos 0\cdot \sin \frac{\pi }{2}\cdot {\cos }^2\left(\frac{\pi }{6}\right)\cdot {\sec }^2\left(\frac{\pi }{4}\right)}{\tan \frac{\pi }{3}+\cot \frac{\pi }{3}}$

Answer 1

From the table $\cos 0=1$
$\sin \frac{\pi }{2}=\sin {90}^o=1$
$\cos \frac{\pi }{6}=\cos {30}^o=\frac{\sqrt{3}}{2}$
$\sec \frac{\pi }{4}=\sec {45}^o=\sqrt{2}$
$\cot \frac{\pi }{3}=\cot {60}^o=\frac{1}{\sqrt{3}}$
$\tan \frac{\pi }{3}=\tan {60}^o=\sqrt{3}$
By substitution, $\frac{1\times 1\times {\left(\frac{\sqrt{3}}{2}\right)}^2\cdot {\left(\sqrt{2}\right)}^2}{\sqrt{3}+\frac{1}{\sqrt{3}}}=\frac{1\times 1\times \frac{3}{4}\times 2}{\frac{3+1}{\sqrt{3}}}=\frac{3}{2}\times \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{8}$