Question

$4ta{n}^{-1}\frac{1}{5}-ta{n}^{-1}\frac{1}{239}$ is equal to

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

Answer 1

The correct option is D $\frac{\pi }{4}$

Since $2ta{n}^{-1}x=ta{n}^{-1}\frac{2x}{1-x^2}$
$\therefore 4ta{n}^{-1}\frac{1}{5}=2\left[2ta{n}^{-1}\frac{1}{5}\right]=2ta{n}^{-1}\frac{\frac{2}{5}}{1-\frac{1}{25}}$
$=2ta{n}^{-1}\frac{10}{24}=ta{n}^{-1}\frac{\frac{20}{24}}{1-\frac{100}{576}}=ta{n}^{-1}\frac{120}{119}$
So, $4ta{n}^{-1}\frac{1}{5}-ta{n}^{-1}\frac{1}{239}=ta{n}^{-1}\frac{120}{119}-ta{n}^{-1}\frac{1}{239}$
$=ta{n}^{-1}\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}.\frac{1}{239}}=ta{n}^{-1}\frac{(120\times 239)-119}{(119\times 239)+120}$
$\Rightarrow ta{n}^{-1}1=\frac{\pi }{4}$.