Question

Given $\frac{x-x{\tan }^2{30}^o}{1+{\tan }^2{30}^o}={\sin }^2{30}^o+4{\cot }^2{45}^o-{\sec }^2{45}^o$. Then the value of $x$

• A. $\frac{3}{2}$
• B. $\frac{5}{2}$
• C. $\frac{9}{2}$
• D. none of these

Answer 1

Answer: (C)

$\frac{9}{2}$

Given,

$\frac{x-x{\tan }^2{30}^o}{1+{\tan }^2{30}^o}={\sin }^2{30}^o+4{\cot }^2{45}^o-{\sec }^2{45}^o$

or, $\frac{x-\frac{x}{3}}{1+\frac{1}{3}}=\frac{1}{4}+4-2$ [ Using the values of $\tan {30}^o,\cot {45}^o,\sin {30}^o,\sec {45}^o$]

or, $\frac{\frac{2x}{3}}{\frac{4}{3}}=\frac{9}{4}$

or, $x=\frac{9}{2}$.