Question

Show that:
$4\sin \frac{\pi }{6}{\sin }^2\frac{\pi }{3}+3\cos \frac{\pi }{3}\tan \frac{\pi }{4}+{\mathrm{cosec}}^2\frac{\pi }{2}=2{\sec }^2\frac{\pi }{4}$

Answer 1

Substituting the value of $\sin \frac{\pi }{6}=\frac{1}{2},\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2},\cos \frac{\pi }{3}=\frac{1}{2},\tan \frac{\pi }{4}=1$ and $\csc \frac{\pi }{2}=1$ we have

$4\sin \frac{\pi }{6}{\sin }^2\frac{\pi }{3}+3\cos \frac{\pi }{3}\tan \frac{\pi }{4}+{\csc }^2\frac{\pi }{2}$

$=4\times \frac{1}{2}\times \frac{3}{4}+3\times \frac{1}{2}\times 1+1$

$=\frac{3}{2}+\frac{3}{2}+1$

$=\frac{6}{2}+1$

$=3+1$

$=4$

$=2\times {\left(\sqrt{2}\right)}^2$

$=2{\sec }^2\frac{\pi }{4}$ since $\sec \frac{\pi }{4}=\sqrt{2}$