Question

Prove that:

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

Answer 1

We have

L.H.S = $2{\text{sin}}^2\frac{\pi }{6}+{\text{cosec}}^2\frac{7\pi }{6}{\text{cos}}^2\frac{\pi }{3}$

Now $\text{cosec}\frac{7\pi }{6}=\text{cosec}(\pi +\frac{\pi }{6}){\text{cos}}^2\frac{\pi }{3}$

= $2\times {\left(\frac{1}{2}\right)}^2+(-2{)}^2\times {\left(\frac{1}{2}\right)}^2$

= $\frac{2\times 1}{4}+\frac{4\times 1}{4}=\frac{6}{4}=\frac{3}{2}$ = R.H.S