Question

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

Answer 1

We have,
$sec\alpha =\frac{Hypotenuse}{Bse}=\frac{5}{4}$
So, we draw a right triangle ABC, right angled at B such that
Hypotenuse = AC = 5 units, Base = 4 units, and $\angle$BAC = $\alpha .$
By Pythagoras theorem, we have
$A{C}^2=A{B}^2+B{C}^2$
$\Rightarrow 5^2=4^2+B{C}^2$
$\Rightarrow B{C}^2=5^2-4^2=9.$
$\Rightarrow BC=\sqrt{9}=3$
$\therefore tan\alpha =\frac{BC}{AB}=\frac{3}{4}$
Now, $\frac{1-tan\alpha }{1+tan\alpha }=\frac{1-\frac{3}{4}}{1+\frac{3}{4}}=\frac{\frac{1}{4}}{\frac{7}{4}}=\frac{1}{7}$
Note : It should be noted that :
$si{n}^2\theta =(sin\theta {)}^2,si{n}^3\theta =(sin\theta {)}^3,co{s}^3\theta =(cos\theta {)}^3$ etc.