Question

cos (tan^{-1} \frac{2}{3})

Explanation:

Step1: Let $\theta=\tan^{-1}\frac{2}{3}$

By the definition of inverse - tangent function, if $\theta = \tan^{-1}\frac{2}{3}$, then $\tan\theta=\frac{2}{3}$ and $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$.

Step2: Use the identity $\tan\theta=\frac{y}{x}$ and $r = \sqrt{x^{2}+y^{2}}$

If $\tan\theta=\frac{y}{x}=\frac{2}{3}$ (where we can assume $y = 2$ and $x = 3$), then by the Pythagorean theorem $r=\sqrt{x^{2}+y^{2}}=\sqrt{3^{2}+2^{2}}=\sqrt{9 + 4}=\sqrt{13}$.

Step3: Use the cosine formula $\cos\theta=\frac{x}{r}$

Since $\cos\theta=\frac{x}{r}$ and $x = 3$, $r=\sqrt{13}$, then $\cos\theta=\frac{3}{\sqrt{13}}=\frac{3\sqrt{13}}{13}$.

Answer:

$\frac{3\sqrt{13}}{13}$