Question

in the right triangle ABC , right angle at B , the ratio of AB to AC is 1 : $\sqrt{2}$ find the values of
$\frac{2tanA}{1+ta{n}^{2}A}$

Answer 1

AB : AC = 1 : $\sqrt{2}\Rightarrow \frac{AB}{AC}=\frac{1}{\sqrt{2}}$
$\therefore$ AB = x and AC = $\sqrt{2}$x, for some x.
By Pythagoras theorem, we have
$A{C}^2=A{B}^2+B{C}^2$
$\Rightarrow (\sqrt{2}x{)}^2=x^2+B{C}^2$
$\Rightarrow B{C}^2=2x^2-x^2=x^2$
$\Rightarrow BC=x$
$\therefore tanA=\frac{BC}{AB}=\frac{x}{x}=1$
we have,
$\frac{2tanA}{1+ta{n}^{2}A}=\frac{2\times 1}{1+1}=\frac{2}{2},$ = $1$