Question

Verify the following equalities.
$4{\cot }^2{45}^{\circ }-{\sec }^2{60}^{\circ }+{\sin }^2{60}^{\circ }+{\cos }^2{60}^{\circ }=1$

Answer 1

$4{\cot }^{2}45°-{\sec }^{2}60°+{\sin }^{2}60°+{\cos }^{2}60°$

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

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

$=-2+1$

$=-1.$

Hence, the answer is $-1.$