Question

If $3ta{n}^{-1}\frac{1}{2+\sqrt{3}}-ta{n}^{-1}\frac{1}{x}=ta{n}^{-1}\frac{1}{3}$, then $x=$

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. 2
• D. 4

Answer 1

The correct option is C

2

Consider
$ta{n}^{-1}\frac{1}{2+\sqrt{3}}=ta{n}^{-1}\frac{2-\sqrt{3}}{4-3}$
$ta{n}^{-1}(2-\sqrt{3})={15}^{\circ }$
$3ta{n}^{-1}\frac{1}{2+\sqrt{3}}=3\left({15}^{\circ }\right)={45}^{\circ }=ta{n}^{-1}1$
Hence the given equation can be rewritten as
$ta{n}^{-1}1-ta{n}^{-1}\frac{1}{3}=ta{n}^{-1}\frac{1}{x}$
$ta{n}^{-1}\frac{1-\frac{1}{3}}{1-\frac{1}{3}}=ta{n}^{-1}\frac{1}{x}$
$ta{n}^{-1}\frac{2}{4}=ta{n}^{-1}\frac{1}{x}$
$\frac{2}{4}=\frac{1}{x}$
$x=2$