Question

2 tan inverse 1 by 5 + 10 inverse 1 by 8 = Sin inverse 4 by 7

Answer 1

Solution :

[tex]2tan^{-1} \frac{1}{5} = tan^{-1} \frac{2( \frac{1}{5} ) }{1-( \frac{1}{5})^2 } = tan^{-1} ( \frac{5}{12} ) \ \ | \ \ | \\ \\ tan^{-1} \frac{1}{8} + 2tan^{-1} \frac{1}{5} = tan^{-1} \frac{1}{8} + tan^{-1} \frac{5}{12} \\ \\ = tan^{-1} \frac{\frac{1}{8} +\frac{5}{12} }{1- \frac{1}{8} \frac{5}{12} } = tan^{-1} \frac{4}{7} \\ \\ Hence \ \ tan^{-1} \frac{1}{8} + 2tan^{-1} \frac{1}{5} = tan^{-1} \frac{4}{7}[/tex]

Only necessary formulas used : 
[tex]tan^{-1} x + tan^{-1} y = tan^{-1} \frac{x+y}{1-xy} \\ \\ 2tan^{-1} x = tan^{-1} \frac{2x}{1-x^2}[/tex]