Question

If $\sec \alpha =\frac{5}{4}$, evaluate $\frac{1-\tan \alpha }{1+\tan \alpha }$

Answer 1

$\sec \alpha =\frac{5}{4}=\frac{Hypotenuse}{Base}$

$Perpendicula{r}^2=5^2-4^2=9$

$Perpendicular=3$

$\tan \alpha =\frac{3}{4}$

Therefore,

$\frac{1-\tan \alpha }{1+\tan \alpha }=\frac{1/4}{7/4}=\frac{1}{7}$