Question

Find $x$ if $x{\sin }^2{60}^0-\frac{1}{2}\sec {60}^0{\tan }^2{30}^0+\frac{4}{3}{\sin }^2{45}^0{\tan }^2{60}^0=0$

• A. $\frac{-20}{9}$
• B. $\frac{2}{9}$
• C. $\frac{20}{9}$
• D. None of these

Answer 1

Answer: (A)

$\frac{-20}{9}$

$x{\sin }^2{60}^0-\frac{1}{2}\sec {60}^0{\tan }^2{30}^0+\frac{4}{3}{\sin }^2{45}^0{\tan }^2{60}^0=0$

or $x.{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{1}{2}\times 2\times {\left(\frac{1}{\sqrt{3}}\right)}^2-\frac{4}{3}\times {\left(\frac{1}{\sqrt{2}}\right)}^2\times {\left(\sqrt{3}\right)}^2$

or $x.\left(\frac{3}{4}\right)=\frac{1}{3}-\frac{4}{3}\times \frac{1}{2}\times 3$

$=\frac{1}{3}-2=-\frac{5}{3}$

$\therefore x=-\frac{5}{3}\times \frac{4}{3}=-\frac{20}{9}$