Question

$3{\tan }^2{30}^{\circ }+\frac{4}{3}{\cos }^2{30}^{\circ }-2{\sin }^2{45}^{\circ }-\frac{1}{3}{\sin }^2{60}^{\circ }$ is equal to__________________.

• A. $\frac{1}{4}$
• B. $\frac{3}{4}$
• C. $1$
• D. $0$

Answer 1

The correct option is B $\frac{3}{4}$

Given

$3{\tan }^{2}30+\frac{4}{3}{\cos }^{2}30-2{\sin }^{2}45-\frac{1}{3}{\sin }^{2}60$

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

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

$=\frac{3}{4}$