Question

Evaluate:
${\tan }^2{30}^{}\cdot \sin {30}^{}+\cos {60}^{}\cdot {\sin }^2{90}^{}\cdot {\tan }^2{60}^{}-2\tan {45}^{}\cdot {\cos }^2{0}^{}\cdot \sin {90}^{}$

Answer 1

${\tan }^{2}30°.\sin 30°+\cos 60°.{\sin }^{2}90°{\tan }^{2}60°-2\tan 45°.{\cos }^{2}0°.\sin 90°$

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

$=\frac{1}{6}+\frac{3}{2}-2$

$=\frac{1+9-12}{6}$

$=\frac{-2}{6}$

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

Hence, the answer is $\frac{-1}{3}.$