Answers: 2
![\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](/tpl/images/0717/0510/11246.png)
![\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}](/tpl/images/0717/0510/96bbe.png)
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