Question

The value of $ta{n}^2{30}^0+4si{n}^2{45}^0+\frac{1}{3}co{s}^2{30}^0$ is

• A. $2\frac{7}{12}$
• B. $1\frac{5}{12}$
• C. $-2\frac{5}{12}$
• D. $-1\frac{5}{12}$

Answer 1

The correct option is A $2\frac{7}{12}$

$ta{n}^2{30}^0+4si{n}^2{45}^0+\frac{1}{3}co{s}^2{30}^0$
$={\left(\frac{1}{\sqrt{3}}\right)}^2+4{\left(\frac{1}{\sqrt{2}}\right)}^2+\frac{1}{3}{\left(\frac{\sqrt{3}}{2}\right)}^2$
$=\frac{1}{3}+\frac{4}{2}+\frac{3}{3\times 4}$
$=\frac{4+24+3}{12}$
$=\frac{31}{12}$
$=2\frac{7}{12}$