How do you find the value for #csc (Tan^-1 (4/5))#?

To find the value of $\csc(\tan^{-1}(4/5))$, we first need to determine the angle whose tangent is $4/5$, then find the cosecant of that angle.

Given that $\tan(\theta) = \frac{4}{5}$, we can find the angle $\theta$ by taking the arctangent (inverse tangent) of $4/5$:

$\theta = \tan^{-1}\left(\frac{4}{5}\right)$

Using a calculator or trigonometric tables, we find that $\theta \approx 38.66^\circ$.

Now, to find $\csc(\theta)$, we recall that $\csc(\theta)$ is the reciprocal of the sine function:

$\csc(\theta) = \frac{1}{\sin(\theta)}$

Since $\sin(\theta)$ is the opposite over the hypotenuse in a right triangle, and we know that the tangent of the angle is $4/5$, we can use the Pythagorean theorem to find the adjacent side:

$a^2 + b^2 = c^2$

$4^2 + 5^2 = c^2$

$16 + 25 = c^2$

$c^2 = 41$

$c = \sqrt{41}$

Therefore, $\csc(\theta) = \frac{\sqrt{41}}{4}$.

So, the value of $\csc(\tan^{-1}(4/5))$ is $\frac{\sqrt{41}}{4}$.