Reference for the trig identities $\cos\phi= \frac{1}{\sqrt{1+\tan^2\phi}}$ and $\sin\phi=\frac{\tan\phi}{\sqrt{1+\tan^2\phi}}$?

Question

What reference has the trigonometric identities below?

$$\begin{align} \cos\phi &= \frac{1}{\sqrt{1 + \tan^2\phi}} \\ \sin\phi &= \frac{\tan\phi}{\sqrt{1 + \tan^2\phi}} \end{align}$$

Your help will be highly appreciated.

Answer 1

Suppose you already know the identity $\sin^2 \phi + \cos^2 \phi = 1$. Then: $$\begin{align*} \frac{1}{\sqrt{1+\tan^2(\phi)}} &= \frac{1}{\sqrt{1+\frac{\sin^2 \phi}{\cos^2\phi}}}\end{align*}\\ =\frac{1}{\sqrt{\frac{\cos^2\phi}{\cos^2\phi}+\frac{\sin^2 \phi}{\cos^2\phi}}}\\ = \frac{1}{\sqrt{\frac{\cos^2\phi + \sin^2 \phi}{\cos^2\phi}}}\\ = \frac{1}{\sqrt{\frac{1}{\cos^2\phi}}}\\ = \cos{\phi} $$

Also, then we have $\frac{\tan \phi}{\sqrt{1+\tan^2(\phi)}} = \tan\phi \cdot\frac{1}{\sqrt{1+\tan^2{\phi}}} = \tan{\phi}\cdot\cos{\phi} = \frac{\sin\phi}{\cos\phi}\cos\phi = \sin\phi.$

Answer 2

I have no reference, but a derivation is that in the first quadrant is $$ \frac{1}{\sqrt{1 + \tan^2(\phi)}} = \frac{1}{\sqrt{1 + \frac{\sin^2(\phi)}{\cos^2(\phi)}}} = \sqrt{\frac{\cos^2(\phi)}{\sin^2(\phi) + \cos^2(\phi)}} = \cos(\phi)$$ and then $$ \frac{\tan(\phi)}{\sqrt{1 + \tan^2(\phi)}} = \tan(\phi) \cdot \frac{1}{\sqrt{1 + \tan^2(\phi)}} = \tan(\phi) \cos(\phi) = \frac{\sin(\phi)}{\cos(\phi)} \cdot \cos(\phi) = \sin(\phi) $$ It's also worth noting that the question as stated isn't careful about signs. I've presumed that to fix this we could fix our domain to the first quadrant, but you could also do the casework, taking the equality in absolute values, to use this in other regions of the unit circle as well.