Question

Verify that ${\cot }^2{60}^o+{\sin }^2{45}^o+{\sin }^2{30}^o+{\cos }^2{90}^o=\frac{13}{12}$

Answer 1

${\cot }^2{60}^0+{\sin }^2{45}^{\circ }+{\sin }^2{30}^{\circ }+{\cos }^2{90}^{\circ }$

$={\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{2}\right)}^2+0$

$=\frac{1}{3}+\frac{1}{2}+\frac{1}{4}$

$=\frac{4+6+3}{12}$

$=\frac{13}{12}$