Question

${\displaystyle 4{\tan }^{-1} \frac{1}{5}-{\tan }^{-1} \frac{1}{239}}$ का मान ज्ञात कीजिए।

Answer 1

${\displaystyle 4{\tan }^{-1} \frac{1}{5}-{\tan }^{-1} \frac{1}{239}}$

= ${\displaystyle 2\left(2{\tan }^{-1} \frac{1}{5}\right)-{\tan }^{-1} \frac{1}{239}}$

= ${\displaystyle 2{\tan }^{-1} \frac{\frac{2}{5}}{1-{\left(\frac{1}{5}\right)}^2}-{\tan }^{-1} \frac{1}{239}}$  .....${\displaystyle (\text{क्योंकि} 2{\tan }^{-1}x={\tan }^{-1} \frac{2x}{1-x^2})}$

= ${\displaystyle 2{\tan }^{-1} \frac{\frac{2}{5}}{\frac{24}{25}}-{\tan }^{-1} \frac{1}{239}}$

= ${\displaystyle 2{\tan }^{-1} \frac{5}{12}-{\tan }^{-1} \frac{1}{239}}$

= ${\displaystyle 2{\tan }^{-1} \frac{\frac{2}{5}}{1-{\left(\frac{1}{5}\right)}^2}-{\tan }^{-1} \frac{1}{239}}$ .....${\displaystyle (\text{क्योंकि} 2{\tan }^{-1}x={\tan }^{-1} \frac{2x}{1-x^2})}$

= ${\displaystyle 2{\tan }^{-1} \frac{\frac{2}{5}}{\frac{24}{25}}-{\tan }^{-1} \frac{1}{239}}$

= ${\displaystyle 2{\tan }^{-1} \frac{5}{12}-{\tan }^{-1} \frac{1}{239}}$

= ${\displaystyle {\tan }^{-1} \frac{2\cdot \frac{5}{12}}{1-{\left(\frac{5}{12}\right)}^2}-{\tan }^{-1} \frac{1}{239}}$  ......${\displaystyle (\text{क्योंकि} 2{\tan }^{-1}x={\tan }^{-1} \frac{2x}{1-x^2})}$

= ${\displaystyle {\tan }^{-1} \frac{144\times 5}{119\times 6}-{\tan }^{-1} \frac{1}{239}}$

= ${\displaystyle {\tan }^{-1} \frac{120}{119}-{\tan }^{-1} \frac{1}{239}}$

= ${\displaystyle {\tan }^{-1} \frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}\cdot \frac{1}{239}}}$  ......${\displaystyle (\text{क्योंकि}{\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1} \frac{x-y}{1+xy})}$

= ${\displaystyle {\tan }^{-1} \frac{120\times 239-119}{119\times 239+120}}$

= ${\displaystyle {\tan }^{-1} \frac{28680-119}{28441+120}}$

= ${\displaystyle {\tan }^{-1} \frac{28561}{28561}}$

= ${\displaystyle {\tan }^{-1}1=\frac{\pi }{4}}$