Question

in the right triangle shown below, one of the interior angles is equal to $\tan^{-1}left(\frac{b}{a}
ight)$. which of the expressions shown is equivalent to $sinleft\tan^{-1}left(\frac{b}{a}
ight)
ight$? $\frac{sqrt{a^{2}+b^{2}}}{b}$ $\frac{b}{sqrt{a^{2}+b^{2}}}$ $\frac{a}{sqrt{a^{2}+b^{2}}}$ $\frac{b}{a}$ $\frac{sqrt{a^{2}+b^{2}}}{a}$

Explanation:

Step1: Let $\theta=\tan^{-1}(\frac{b}{a})$.

This means $\tan\theta = \frac{b}{a}$ in a right - triangle. By the definition of tangent, if $\tan\theta=\frac{b}{a}$, we can consider a right - triangle where the opposite side to the angle $\theta$ is $b$ and the adjacent side is $a$.

Step2: Find the hypotenuse.

Using the Pythagorean theorem $c^{2}=a^{2}+b^{2}$, where $c$ is the hypotenuse. So, $c = \sqrt{a^{2}+b^{2}}$.

Step3: Find $\sin\theta$.

By the definition of sine in a right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. Since $\theta=\tan^{-1}(\frac{b}{a})$, and in our right - triangle the opposite side to $\theta$ is $b$ and the hypotenuse is $\sqrt{a^{2}+b^{2}}$, then $\sin[\tan^{-1}(\frac{b}{a})]=\frac{b}{\sqrt{a^{2}+b^{2}}}$.

Answer:

$\frac{b}{\sqrt{a^{2}+b^{2}}}$ (the second option in the multiple - choice list)