Question

in a right ∆ABC, right angled at B and AB:AC=1:√5. Find the value of 2tanA/1+tan²A​

Answer 1

• We are given that,

In right triangle ABC, the ratio $AB : AC = 1 : \sqrt{2}$.

• So, from the figure below, we get that,

$\cos A=\frac{AB}{AC}=\frac{1}{\sqrt{2}}$

• i.e. $\cos A=\frac{1}{\sqrt{2}}$

$\cos A=\cos 45$

• i.e. A = 45°

So, we have,

$\frac{2\tan A}{1+tan^2A}=\frac{2\tan 45}{1+tan^245}$

i.e. $\frac{2\tan A}{1+tan^2A}=\frac{2\times 1}{1+1^2}$

i.e. $\frac{2\tan A}{1+tan^2A}=\frac{2}{2}$

i.e. $\frac{2\tan A}{1+tan^2A}=1$

Hence, the value of $\frac{2\tan A}{1+tan^2A}$ is 1.