Rewrite $\tan \left(\cos ^{-1} v\right)$ as an algebraic expression in $v$.
\[
\tan \left(\cos ^{-1} v\right)=
\]

Answer

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Answer

$\boxed{\tan (\cos ^{-1} v) = \frac{\sqrt{1 - v^2}}{v}}$ is the final answer

Steps

Step 1 ：Given the expression $\tan (\cos ^{-1} v)$

Step 2 ：Recall the Pythagorean identity $1 = \sin^2(\theta) + \cos^2(\theta)$

Step 3 ：Since $\cos(\theta) = v$, we can solve for $\sin(\theta)$ to get $\sin(\theta) = \sqrt{1 - v^2}$

Step 4 ：Then, use the definition of $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ to rewrite the expression in terms of v

Step 5 ：Substitute $\sin(\theta)$ and $\cos(\theta)$ into the equation to get $\tan (\cos ^{-1} v) = \frac{\sqrt{1 - v^2}}{v}$

Step 6 ：$\boxed{\tan (\cos ^{-1} v) = \frac{\sqrt{1 - v^2}}{v}}$ is the final answer