rewrite \\( \\cos \\left( \\tan ^ { - 1 } 4 v \\right) \\) as an algebraic expression in \\( v \\).\n\\…

Explanation:

Step1: Let (\theta=\tan^{- 1}(4v))

By the definition of inverse tangent function, (\tan\theta = 4v=\frac{4v}{1}), where (\theta\in(-\frac{\pi}{2},\frac{\pi}{2})). Consider a right - triangle where the opposite side (y = 4v) and the adjacent side (x = 1).

Step2: Find the hypotenuse (r)

Using the Pythagorean theorem (r=\sqrt{x^{2}+y^{2}}). Substitute (x = 1) and (y = 4v) into the formula, we get (r=\sqrt{1+(4v)^{2}}=\sqrt{1 + 16v^{2}}).

Step3: Find (\cos\theta)

By the definition of cosine function (\cos\theta=\frac{x}{r}). Since (x = 1) and (r=\sqrt{1 + 16v^{2}}), and (\theta=\tan^{-1}(4v)), then (\cos(\tan^{-1}(4v))=\frac{1}{\sqrt{1 + 16v^{2}}}).

Answer:

(\frac{1}{\sqrt{1 + 16v^{2}}})