Question

$si{n}^2{30}^{\circ }+co{s}^{2}60+{\tan }^2{45}^{\circ }+se{c}^2{60}^{\circ }-co{\sec }^2{30}^{\circ }=\frac{3}{2}$

Answer 1

$=si{n}^2{30}^{\circ }+co{s}^2{60}^{\circ }+ta{n}^2{45}^{\circ }+se{c}^2{60}^{\circ }-cose{c}^2{30}^{\circ }$
$=si{n}^2{30}^{\circ }+si{n}^2{30}^{\circ }+1+cose{c}^2{30}^{\circ }-cose{c}^2{30}^{\circ }$
$=2si{n}^2{30}^{\circ }+1$
$=2\ast \frac{1}{4}+1$
$=\frac{1}{2}+1=\frac{3}{2}$