Question

The value of $\frac{{\tan }^2{60}^o-2{\tan }^2{45}^o+{\sec }^2{30}^o}{3{\sin }^2{45}^o\sin {90}^o+{\cos }^2{60}^o{\cos }^3{0}^o}$ is

• A. $\frac{49}{12}$
• B. $\frac{7}{3}$
• C. $\frac{14}{9}$
• D. $\frac{4}{3}$

Answer 1

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

$\frac{{\tan }^2{60}^o-2{\tan }^2{45}^o+{\sec }^2{30}^o}{3{\sin }^2{45}^o\sin {90}^o+{\cos }^2{60}^o{\cos }^3{0}^o}$

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