\[2\tan^{-1}\frac35=\tan^{-1}\left(\frac{2\cdot\frac35}{1-\frac{9}{25}}\right)=\tan^{-1}\frac{15}{8}.\]
Let θ=sin−1135: tanθ=125.
\[\tan\left(\tan^{-1}\frac{15}{8}+\tan^{-1}\frac{5}{12}\right)=\frac{\frac{15}{8}+\frac{5}{12}}{1-\frac{15}{8}\cdot\frac{5}{12}}=\frac{\frac{55}{24}}{\frac{21}{96}}=\frac{55}{24}\cdot\frac{96}{21}=\frac{220}{21}.\]