Question

$\frac{{\tan }^2{60}^0+4{\sin }^2{45}^0+3{\sec }^2{30}^0+5{\cos }^2{90}^0}{\csc {30}^0+\sec {60}^0-{\cot }^2{30}^0}=$

• A. $\frac{9}{7}$
• B. $9$
• C. $\frac{7}{9}$
• D. $6$

Answer 1

The correct option is B $9$

Given that: $\frac{{\tan }^2{60}^0+4{\sin }^2{45}^0+3{\sec }^2{30}^0+5{\cos }^2{90}^0}{\csc {30}^0+\sec {60}^0-{\cot }^2{30}^0}$

$=\frac{{\left(\sqrt{3}\right)}^2+4\times {\left(\frac{1}{\sqrt{2}}\right)}^2+3\times {\left(\frac{2}{\sqrt{3}}\right)}^2+5\times 0}{2+2-{\left(\sqrt{3}\right)}^2}$

$=\frac{3+4\times \frac{1}{2}+3\times \frac{4}{3}+0}{4-3}$

$=\frac{3+2+4+0}{1}$

$=9$