Question

If $\tan (\alpha -\beta )=\frac{7}{24}$ and $\tan \alpha =\frac{4}{3}$ where $\alpha$ and $\beta$ are in the first quadrant prove that $\alpha +\beta =\pi /2$

Answer 1

$tan(\alpha -\beta )=\frac{7}{24}$

$\frac{tan\alpha -tan\beta }{1+tan\alpha tan\beta }=\frac{7}{24}$

$\frac{\frac{4}{3}-tan\beta }{1+\frac{4}{3}tan\beta }=\frac{7}{24}$

by solving we get $tan\beta =\frac{3}{4}$

now $tan(\alpha +\beta )=\frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta }$

$=\frac{\frac{4}{3}+\frac{3}{4}}{0}=doesnotexit$

therefore $(\alpha +\beta )=\frac{\pi }{2}$