Question

If $ta{n}^{-1}\left(\frac{1}{4}\right)+ta{n}^{-1}=\left(\frac{2}{9}\right)=\frac{1}{2}co{s}^{-1}x,$ then x is equal to

• A. $\frac{1}{2}$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

The correct option is C $\frac{3}{5}$

Given $\frac{1}{2}co{s}^{-1}x=ta{n}^{-1}\left(\frac{1}{4}\right)+ta{n}^{-1}\left(\frac{2}{9}\right)$
$\Rightarrow co{s}^{-1}\left(x\right)=2ta{n}^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}.\frac{2}{9}}\right)\Rightarrow co{s}^{-1}x=2ta{n}^{-1}\frac{1}{2}$
$\Rightarrow x=\frac{1-ta{n}^2\left(ta{n}^{-1}\left(\frac{1}{2}\right)\right)}{1+ta{n}^2\left(ta{n}^{-1}\left(\frac{1}{2}\right)\right)}=\frac{1-\frac{1}{4}}{1+\frac{1}{4}}=\frac{3}{5}$