Question

Prove that : $2{\sin }^2\frac{\pi }{6}+{\text{cosec}}^2\frac{7\pi }{6}{\cos }^2\frac{\pi }{3}=\frac{3}{2}$

Answer 1

$2{\sin }^2\frac{\pi }{6}+cose{c}^2\frac{7\pi }{6}{\cos }^2\frac{\pi }{3}$

$=2{\left(\frac{1}{2}\right)}^2+cose{c}^2(\pi +\frac{\pi }{6})\cdot {\left(\frac{1}{2}\right)}^2$

$=\frac{1}{2}+{(-cosec\frac{\pi }{6})}^2\cdot \frac{1}{4}$

$=\frac{1}{2}+(-2{)}^2\cdot \frac{1}{4}$

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

$=\frac{3}{2}$