Using the Pythagorean Identity prove (sinTheta )/(sqrt (1-sin^2)Theta )=tanTheta In which quadrants is this equation true?

To prove the given equation using the Pythagorean Identity, we can start by recalling the Pythagorean Identity:<br /><br />\[ \sin^2\Theta + \cos^2\Theta = 1 \]<br /><br />Now, let's manipulate this identity to find a relationship between \(\sin\Theta\) and \(\cos\Theta\):<br /><br />\[ \cos^2\Theta = 1 - \sin^2\Theta \]<br /><br />Taking the square root of both sides, we get:<br /><br />\[ \cos\Theta = \sqrt{1 - \sin^2\Theta} \]<br /><br />Now, let's substitute this expression for \(\cos\Theta\) into the given equation:<br /><br />\[ \frac{\sin\Theta}{\sqrt{1 - \sin^2\Theta}} = \tan\Theta \]<br /><br />We know that \(\tan\Theta = \frac{\sin\Theta}{\cos\Theta}\). So, we can rewrite the equation as:<br /><br />\[ \frac{\sin\Theta}{\sqrt{1 - \sin^2\Theta}} = \frac{\sin\Theta}{\cos\Theta} \]<br /><br />Since \(\cos\Theta = \sqrt{1 - \sin^2\Theta}\), we can simplify the equation further:<br /><br />\[ \frac{\sin\Theta}{\sqrt{1 - \sin^2\Theta}} = \frac{\sin\Theta}{\sqrt{1 - \sin^2\Theta}} \]<br /><br />This equation is true for all values of \(\Theta\) where \(\cos\Theta\) is defined. However, we need to consider the domain of the original equation, which is \(\sqrt{1 - \sin^2\Theta}\). The square root function is only defined for non-negative values, so we need to ensure that \(1 - \sin^2\Theta \geq 0\).<br /><br />This inequality is true for all values of \(\Theta\) where \(\sin\Theta\) is defined. Therefore, the equation is true for all values of \(\Theta\) where \(\sin\Theta\) is defined.<br /><br />In terms of the quadrants, the equation is true in all four quadrants of the unit circle, as long as \(\sin\Theta\) is defined.