Question

I need help with this question for my Precalculus class

Answer 1

If
`x` is positive, then
`\tan^(-1)(x)` is between 0 and π/2, which means
`\cos(\tan^(-1)(x))` is between 1 and 0.

Let
`\theta = \tan^(-1)(x)`. Then
`\tan(\theta) = x`, and it follows from the Pythagorean identity that

`\sin^2(\theta) + \cos^2(\theta) = 1 \implies \tan^2(\theta) + 1 = \sec^2(\theta) \\\\ ~~~~ \implies \sec(\theta) = √(1 + \tan^2(\theta)) \\\\ ~~~~ \implies \cos(\theta) = \frac1{√(1+\tan^2(\theta))} \\\\ ~~~~ \implies \cos(\tan^(-1)(x)) = \boxed{\frac1{√(1+x^2)}}`

We took the positive square root when solving for
`\sec(\theta)` because we know
`\cos(\theta)` is non-negative, between 0 and 1.