The value of frac{sin^2 theta}{sin theta - co | vedclass

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(D) Let the expression be $E = \frac{\sin^2 	heta}{\sin 	heta - \cos 	heta} - \frac{\sin 	heta + \cos 	heta}{	an^2 	heta - 1}$.First,simplify t

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lies between $-\sqrt{2}$ and $\sqrt{2}$

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(D) Let the expression be $E = \frac{\sin^2 	heta}{\sin 	heta - \cos 	heta} - \frac{\sin 	heta + \cos 	heta}{	an^2 	heta - 1}$.First,simplify the second term: $	an^2 	heta - 1 = \frac{\sin^2 	heta}{\cos^2 	heta} - 1 = \frac{\sin^2 	heta - \cos^2 	heta}{\cos^2 	heta} = \frac{(\sin 	heta - \cos 	heta)(\sin 	heta + \cos 	heta)}{\cos^2 	heta}$.Substituting this into the second term: $\frac{\sin 	heta + \cos 	heta}{\frac{(\sin 	heta - \cos 	heta)(\sin 	heta + \cos 	heta)}{\cos^2 	heta}} = \frac{(\sin 	heta + \cos 	heta) \cdot \cos^2 	heta}{(\sin 	heta - \cos 	heta)(\sin 	heta + \cos 	heta)} = \frac{\cos^2 	heta}{\sin 	heta - \cos 	heta}$.Now,substitute back into $E$: $E = \frac{\sin^2 	heta}{\sin 	heta - \cos 	heta} - \frac{\cos^2 	heta}{\sin 	heta - \cos 	heta} = \frac{\sin^2 	heta - \cos^2 	heta}{\sin 	heta - \cos 	heta}$.Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $E = \frac{(\sin 	heta - \cos 	heta)(\sin 	heta + \cos 	heta)}{\sin 	heta - \cos 	heta} = \sin 	heta + \cos 	heta$.We know that $\sin 	heta + \cos 	heta = \sqrt{2} \sin(	heta + 45^\circ)$.The range of $\sin(	heta + 45^\circ)$ is $[-1, 1]$,so the range of $\sqrt{2} \sin(	heta + 45^\circ)$ is $[-\sqrt{2}, \sqrt{2}]$.Thus,the value lies between $-\sqrt{2}$ and $\sqrt{2}$.

The value of $\frac{\sin^2 \theta}{\sin \theta - \cos \theta} - \frac{\sin \theta + \cos \theta}{\tan^2 \theta - 1}$ for all permissible values of $\theta$ is:

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