Question

Prove ${\sin }^2{30}^o+{\cos }^2{60}^o+{\tan }^2{45}^o+{\sec }^2{60}^o-{\csc }^2{30}^o=\frac{3}{2}$

Answer 1

${\sin }^2{30}^{\circ }+{\cos }^2{60}^{\circ }+{\tan }^2{45}^{\circ }+{\sec }^2{60}^{\circ }-\cos e{c}^2{30}^{\circ }$

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

$=\frac{2}{4}+1+4-4$

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

$=\frac{3}{2}$