Question

Given that tan A = $\frac{1}{\sqrt{5}}$, what is the value of $\frac{{cos}^{2}A-{2sin}^{2}A}{{cos}^{2}A+2{sin}^{2}A}$ ?

Answer 1

$tanA=\frac{1}{\sqrt{5}}$

Therefore,

$cosA=\frac{\sqrt{5}}{\sqrt{6}}⟹co{s}^{2}A=\frac{5}{6}$

$sinA=\frac{1}{\sqrt{6}}⟹si{n}^{2}A=\frac{1}{6}$

Hence,

$\frac{co{s}^{2}A-2si{n}^{2}A}{co{s}^{2}A+2si{n}^{2}A}=\frac{\frac{5}{6}-\frac{2}{6}}{\frac{5}{6}+\frac{2}{6}}=\frac{3}{7}$