Question

Find the value of:

${\sin }^2{30}^0{\cos }^2{45}^0+4{\tan }^2{30}^0+\frac{1}{2}{\sin }^2{90}^0+\frac{1}{2}{\cot }^2{60}^0$

• A. $2$
• B. $3$
• C. $4$
• D. $\frac{17}{8}$

Answer 1

The correct option is D $\frac{17}{8}$

${\sin }^2{30}^0{\cos }^2{45}^0+4{\tan }^2{30}^0+\frac{1}{2}{\sin }^2{90}^0+\frac{1}{2}{\cot }^2{60}^0$

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

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

$=\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{6}$

$=\frac{3+32+12+4}{24}$

$=\frac{17}{8}$