Question

If $tan\left(2{tan}^{-1}\right(\frac{1}{5})-\frac{\pi }{4})=-\frac{\lambda }{17}$, then $\lambda$ is equal to

Answer 1

$\tan (2{\text{tan}}^{-1}(\frac{1}{5}))=\frac{2\tan \left({\text{tan}}^{-1}\frac{1}{5}\right)}{1-{\tan }^2\left({\text{tan}}^{-1}\frac{1}{5}\right)}=\frac{\frac{2}{5}}{1-\frac{1}{25}}=\frac{10}{24}=\frac{5}{12}$

$\tan \left(2{\text{tan}}^{-1}\right(\frac{1}{5})-\frac{\pi }{4})=\frac{\tan \left(2{\text{tan}}^{-1}\frac{1}{5}\right)-\tan \frac{\pi }{4}}{1+\tan \left(2{\text{tan}}^{-1}\right(\frac{1}{5}\left)\right)\tan \frac{\pi }{4}}=\frac{\frac{5}{12}-1}{1+\frac{5}{12}}=-\frac{7}{17}$

$⟹-\frac{\lambda }{17}=-\frac{7}{17}$

$⟹\lambda =7$