give the exact value of the expression without using a calculator. cos (tan^(-1)(-3)) cos (tan^(-1)(-3)) =…

Explanation:

Step1: Let $\theta=\tan^{-1}(-3)$

This means $\tan\theta=-3=\frac{-3}{1}$, and $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$.

Step2: Consider a right - triangle

If $\tan\theta=\frac{y}{x}=-3$, we can let $y = - 3$ and $x = 1$. Then, by the Pythagorean theorem $r=\sqrt{x^{2}+y^{2}}=\sqrt{1^{2}+(-3)^{2}}=\sqrt{1 + 9}=\sqrt{10}$.

Step3: Find $\cos\theta$

Since $\cos\theta=\frac{x}{r}$, substituting $x = 1$ and $r=\sqrt{10}$, we get $\cos\theta=\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}$.

Answer:

$\frac{\sqrt{10}}{10}$