PROVE √(1-〖cos〗^2 θ)/cosθ=tanθ

[tex]\dfrac{\sqrt{1-\cos^2\theta}}{\cos\theta}=\tan\theta\\\\\text{We know:}\\(1)\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\(2)\tan x=\dfrac{\sin x}{\cos x}\\\\\text{For}\ \sqrt{1-\cos^2\theta}\ \text{use}\ (1):\\\\L_S=\dfrac{\sqrt{1-\cos^2\theta}}{\cos\theta}=\dfrac{\sqrt{\sin^2\theta}}{\cos\theta}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta\ \ \ _{used (2)}\\\\R_S=\tan\theta\\\\L_S=R_S[/tex]

[tex]\text{Of course, this equality is not true for everyone values of}\ \theta.\\\\1^o.\ \cos\theta\neq0\to\theta\neq\dfrac{\pi}{2}+k\pi,\ k\in\mathbb{Z}\\\\2^o.\ \sqrt{\sin^2\theta}=|\sin\theta|,\ \text{therefore}\\\\\text{for}\ \theta\in[2k\pi;\ (2k+1)\pi],\ \sin\theta\geq0\to|\sin\theta|=\sin\theta\\\\\text{therefore}\ \dfrac{|\sin\theta|}{\cos\theta}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta[/tex]

[tex]\text{for}\ \theta\in\bigg((2k-1)\pi;\ 2k\pi\bigg),\ \sin\theta<0\to|\sin\theta|=-\sin\theta\\\\\text{therefore}\ \dfrac{|\sin\theta|}{\cos\theta}=\dfrac{-\sin\theta}{\cos\theta}=-\tan\theta\neq\tan\theta[/tex]

[tex]\text{Conclusion:}\\\\\text{This equality is true for}\ \theta\in\left[2k\pi;\ \dfrac{\pi}{2}+2k\pi\right)\cup\left(\dfrac{\pi}{2}+2k\pi;\ \pi+2k\pi\right],\ k\in\mathbb{Z}[/tex]