Question

value of $\frac{{\tan }^2{60}^{\circ }-2{\tan }^2{45}^{\circ }+{\sec }^2{0}^{\circ }}{3{\sin }^2{45}^{\circ }\sin {90}^{\circ }+{\cos }^2{60}^{\circ }co{s}^3{0}^{\circ }}$

• A. $\frac{49}{12}$
• B. $\frac{7}{3}$
• C. $\frac{14}{9}$
• D. $\frac{8}{7}$

Answer 1

Answer: (D)

$\frac{8}{7}$

Consider the given expression,

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

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

$\Rightarrow \frac{3-2+1}{\frac{3}{2}+\frac{1}{4}}$

$\Rightarrow \frac{2}{\frac{6+1}{4}}$

$\Rightarrow \frac{8}{7}$

Hence, this is the answer.