7. evaluate each expression. assume that all angles are in quadrant i.\na. ( \tan left( cos ^ { - 1 } \frac…

Explanation:

Step1: Let (\theta=\cos^{-1}\frac{2}{3})

By the definition of inverse cosine, (\cos\theta=\frac{2}{3}). In a right - triangle (since (\theta) is in quadrant I), if the adjacent side (x = 2) and the hypotenuse (r=3), then by the Pythagorean theorem (y=\sqrt{r^{2}-x^{2}}=\sqrt{9 - 4}=\sqrt{5}). And (\tan\theta=\frac{y}{x}). So, (\tan(\cos^{-1}\frac{2}{3})=\frac{\sqrt{5}}{2}).

Step2: Let (\alpha=\arctan\frac{5}{4})

By the definition of inverse tangent, (\tan\alpha=\frac{5}{4}). In a right - triangle (since (\alpha) is in quadrant I), if the opposite side (y = 5) and the adjacent side (x = 4), then the hypotenuse (r=\sqrt{x^{2}+y^{2}}=\sqrt{16 + 25}=\sqrt{41}). And (\sin\alpha=\frac{y}{r}). So, (\sin(\arctan\frac{5}{4})=\frac{5}{\sqrt{41}}=\frac{5\sqrt{41}}{41}).

Step3: Use the property of inverse cosine function

For (y = \cos^{-1}x), the domain of (x) is ([-1,1]) and (\cos(\cos^{-1}x)=x) when (x\in[-1,1]). Since (\frac{2}{3}\in[-1,1]), then (\cos(\cos^{-1}(\frac{2}{3}))=\frac{2}{3}).

Answer:

a. (\frac{\sqrt{5}}{2}) b. (\frac{5\sqrt{41}}{41}) c. (\frac{2}{3})