The value of the expression frac{1+sin 2 alph | vedclass

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Question

(D) Let the expression be $E$. $E = \frac{1+\sin 2 \alpha}{\cos 2 \alpha 	an (\alpha - \frac{3\pi}{4})} - \frac{1}{4} \sin 2 \alpha [\cot \frac{\alph

Vedclass · Accepted Answer

$\sin ^2 \alpha$

Vedclass · Answer

(D) Let the expression be $E$. $E = \frac{1+\sin 2 \alpha}{\cos 2 \alpha 	an (\alpha - \frac{3\pi}{4})} - \frac{1}{4} \sin 2 \alpha [\cot \frac{\alpha}{2} - 	an \frac{\alpha}{2}]$ Using $	an(\alpha - \frac{3\pi}{4}) = \frac{	an \alpha - (-1)}{1 + 	an \alpha (-1)} = \frac{	an \alpha + 1}{1 - 	an \alpha} = \frac{\sin \alpha + \cos \alpha}{\cos \alpha - \sin \alpha}$. Also,$\cot \frac{\alpha}{2} - 	an \frac{\alpha}{2} = \frac{\cos^2 \frac{\alpha}{2} - \sin^2 \frac{\alpha}{2}}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = \frac{\cos \alpha}{\frac{1}{2} \sin \alpha} = 2 \cot \alpha$. Substituting these: $E = \frac{(\cos \alpha + \sin \alpha)^2}{(\cos^2 \alpha - \sin^2 \alpha) \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha}} - \frac{1}{4} (2 \sin \alpha \cos \alpha) (2 \cot \alpha)$ $E = \frac{(\cos \alpha + \sin \alpha)^2}{(\cos \alpha - \sin \alpha)(\cos \alpha + \sin \alpha) \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha}} - \sin^2 \alpha$ $E = \frac{(\cos \alpha + \sin \alpha)^2}{(\cos \alpha + \sin \alpha)^2} - \sin^2 \alpha$ $E = 1 - \sin^2 \alpha = \cos^2 \alpha$. Wait,re-evaluating the simplification: $E = 1 - \sin^2 \alpha = \cos^2 \alpha$. Given the options,let's re-check the identity $\cot(\frac{3\pi}{2} + \frac{\alpha}{2}) = -	an \frac{\alpha}{2}$. Correcting the final step: $1 - \sin^2 \alpha = \cos^2 \alpha$. If the intended answer is $\sin^2 \alpha$,there might be a sign error in the problem statement or options. Based on standard trigonometric identities,the result is $\cos^2 \alpha$.

The value of the expression $\frac{1+\sin 2 \alpha}{\cos (2 \alpha-2 \pi) \tan \left(\alpha-\frac{3 \pi}{4}\right)} - \frac{1}{4} \sin 2 \alpha \left[\cot \frac{\alpha}{2}+\cot \left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)\right]$ is:

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