Question

If $\tan A-\frac{1}{\tan A}=\sqrt{2}$, then find the value of ${\tan }^{2}A+\frac{1}{{\tan }^{2}A}$.

Answer 1

$tanA=\frac{1}{tanA}=\sqrt{2}$ (Squaring both sides)

$ta{n}^{2}A+\frac{1}{tanA}-2=2ta{n}^{2}A+\frac{1}{ta{n}^{2}A}=4$