Question

$(3{\cos }^2{30}^0+{\sec }^2{30}^0+2\cos 0^0+3\sin {90}^0-{\tan }^2{60}^0)$ is equal to

• A. $\frac{65}{12}$
• B. $\frac{67}{12}$
• C. $\frac{69}{12}$
• D. None of these

Answer 1

Answer: (B)

$\frac{67}{12}$

$3{\cos }^2{30}^0+{\sec }^2{30}^0+2\cos 0^0+3\sin {90}^0-{\tan }^2{60}^0$

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

Substitute the values from the table, we get

$=\frac{9}{4}+\frac{4}{3}+2+3-3$

$=\frac{67}{12}$