Question

Show that
a) $2\left({\cos }^2{45}^{\circ }+{\tan }^2{60}^{\circ }\right)-6\left({\sin }^2{45}^{\circ }-{\tan }^2{30}^{\circ }\right)=6$
b) $2\left({\cos }^4{60}^{\circ }+{\sin }^4{30}^{\circ }\right)-\left({\tan }^2{60}^{\circ }+{\cos }^2{45}^{\circ }\right)+3{\sec }^2{30}^{\circ }=\frac{3}{4}$

Answer 1

a)$2({\cos }^2{45}^{\circ }+{\tan }^2{60}^{\circ })-6({\sin }^2{45}^{\circ }-{\tan }^2{30}^{\circ })$

$=2[{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\sqrt{3}\right)}^2]-6[{\left(\frac{1}{\sqrt{2}}\right)}^2-{\left(\frac{1}{\sqrt{3}}\right)}^2]=2[\frac{1}{2}+3]-6[\frac{1}{2}-\frac{1}{3}]=2\left[\frac{7}{2}\right]-6\left[\frac{3-2}{6}\right]=7-1=6$

b)$2({\cos }^4{60}^{\circ }+{\sin }^4{60}^{\circ })-({\tan }^2{60}^{\circ }+{\cos }^2{45}^{\circ })+3{\sec }^2{30}^{\circ }$

$=2[{\left(\frac{1}{2}\right)}^4+{\left(\frac{1}{2}\right)}^4]-[{\left(\sqrt{3}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2]+3\times \frac{4}{3}=2[\frac{1}{16}+\frac{1}{16}]-[3+\frac{1}{2}]+4=2\times \frac{2}{16}+4-\frac{7}{2}=\frac{1}{4}+4-\frac{7}{2}=\frac{1+16-14}{4}=\frac{3}{4}$