Question

Find the value of ${\cos }^2{60}^{\circ }+{\sec }^2{30}^{\circ }+{\tan }^2{45}^{\circ }$ ?

• A. $\frac{12}{31}$
• B. $\frac{31}{12}$
• C. $0$
• D. $\frac{11}{\sqrt{3}}$

Answer 1

The correct option is B $\frac{31}{12}$

The value of $\cos 60°$ is $\frac{1}{2}$
The value of $\sec 30°$ is $\frac{2}{\sqrt{3}}$
The value of $\tan 45°$ is $1$
${\cos }^2{60}^{\circ }+{\sec }^2{30}^{\circ }+{\tan }^2{45}^{\circ }=$
$(\frac{1}{2}{)}^2+(\frac{2}{\sqrt{3}}{)}^2+(1{)}^2$
$=\frac{1}{4}+\frac{4}{3}+1=\frac{3+16+12}{12}=\frac{31}{12}$
So, option B is the correct option.