Question

Verify the following equalities.
${\tan }^2{60}^{\circ }-2{\tan }^2{45}^{\circ }-{\cot }^2{30}^{\circ }+2{\sin }^2{30}^{\circ }+\frac{3}{4}cose{c}^2{45}^{\circ }=0$

Answer 1

${\tan }^{2}60°-2{\tan }^{2}45°-{\cot }^{2}30°+2{\sin }^{2}30°+\frac{3}{4}cose{c}^{2}45°$

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

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

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

$=-2+2$

$=0.$

Hence, the answer is $0.$