Question

Show that :
2 (${\cos }^2{45}^{\circ }+{\tan }^2{60}^{\circ }$) - 6 (${\sin }^2{45}^{\circ }-{\tan }^2{30}^{\circ }$) = 6

Answer 1

We have,

2 ($co{s}^2{45}^{\circ }+ta{n}^2{60}^{\circ }$) - 6 ($si{n}^2{45}^{\circ }-ta{n}^2{30}^{\circ }$)

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

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