Question

In a right $\Delta ABC$, right angled at $B$, the ratio of $AB$ to $AC$ is $1:\sqrt{2}$, then $\frac{2\tan A}{1-ta{n}^{2}A}$ is equal to

• A. $2$
• B. $1$
• C. $3$
• D. $undefined$

Answer 1

Answer: (D)

$undefined$

$AB:AC=1:\sqrt{2}$

$AB=kAC=\sqrt{2}k$

where $k=$ constant of proportionality

$\therefore B{C}^2=A{C}^2-A{B}^2=3k^2-k^2=k^2$

$\therefore BC=k$

$\tan A=\frac{BC}{AB}=\frac{k}{k}=1$

$\frac{2\tan A}{1-{\tan }^{2}A}=\frac{2\times 1}{1-1}=\frac{2}{0}=$ under fined