Question

Show that :
2 (${\cos }^4{60}^{\circ }+{\sin }^4{30}^{\circ }$) - (${\tan }^2{60}^{\circ }+{\cot }^2{45}^{\circ }$) + 3 ${\sec }^2{30}^{\circ }=\frac{1}{4}$

Answer 1

$2(co{s}^4{60}^{\circ }+si{n}^4{30}^{\circ })-(ta{n}^2{60}^{\circ }+co{t}^2{45}^{\circ })+3se{c}^2{30}^{\circ }$

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

$=(\frac{1}{16}+\frac{1}{16})-(3+1)+3\times \frac{4}{3}=2\times \frac{1}{8}-4+4=\frac{1}{4}$