Question

${\mathrm{cosec}}^2{45}^{\circ }{\sec }^2{30}^{\circ }\left({\sin }^3{30}^{\circ }+4{\cot }^2{45}^{\circ }-{\sec }^2{60}^{\circ }\right)=\frac{1}{3}.$

Answer 1

Solve L.H.S.

${\mathrm{cosec}}^2{45}^{\circ }{\sec }^2{30}^{\circ }\left({\sin }^3{30}^{\circ }+4{\cot }^2{45}^{\circ }-{\sec }^2{60}^{\circ }\right)$

$={\left(\sqrt{2}\right)}^2.{\left(\frac{2}{\sqrt{3}}\right)}^2\left\{{\left(\frac{1}{2}\right)}^3+4{\left(1\right)}^2-{\left(2\right)}^2\right\}$

$=2\times \frac{4}{3}\left\{\frac{1}{8}+4-4\right\}$

$=\frac{8}{3}\times \frac{1}{8}=\frac{1}{3}$

Hence, L.H.S = R.H.S