Question

In a right triangle ABC, right-angled at B, if tan A = 1, then what equals 2 sin A cos A?

Answer 1

Given, tan A = 1

Now, from the figure, $tanA=\frac{BC}{AB}$
$1=\frac{BC}{AB}$
$\therefore$ $AB=BC$



By pythagoras theorem,
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=B{C}^2+B{C}^2$
$A{C}^2=2B{C}^2$....... (i)
$AC=\sqrt{2}BC=\sqrt{2}AB$

Now,
$SinA=\frac{BC}{AC}$ and $CosA=\frac{AB}{AC}$

So,
$2SinACosA$
$=2\times \frac{BC}{AC}$$\times \frac{AB}{AC}$
$=2\times \frac{B{C}^2}{A{C}^2}$
$=2\times \frac{B{C}^2}{2B{C}^2}$ [from (i)]
$=\frac{2}{2}$
$=1$