1+sin2 αcos(2α-2π) tanα-3π4 -14sin2 α cot α2+cot3π2+α2=

1+sin2 αcos(2α-2π) tanα-3π4 -14sin2 α cot α2+cot3π2+α2 =1+sin2 αcos2 α1+tan α1-tan α-14sin2 α cot α2-tan α2 =1+sin2 α1-tan2α1+tan2α 1+tanα1-tanα -14 2sin α cos α 2cot α =1+sin2 α(1+tan α)21+tan2 α -cos2 α =1+2tan α1+tan2α(1+tan α)21+tan2α-cos2α =1+tan2α+2tan α1+tan2α(1+tanα)21+tan2α-cos2α =1-cos2α =sin2α

1+sin2 αcos(2α-2π) tanα-3π4 -14sin2 α cot α2+cot3π2+α2=

1+sin2 αcos(2α-2π) tanα-3π4 -14sin2 α cot α2+cot3π2+α2 =1+sin2 αcos2 α1+tan α1-tan α-14sin2 α cot α2-tan α2 =1+sin2 α1-tan2α1+tan2α 1+tanα1-tanα -14 2sin α cos α 2cot α =1+sin2 α(1+tan α)21+tan2 α -cos2 α =1+2tan α1+tan2α(1+tan α)21+tan2α-cos2α =1+tan2α+2tan α1+tan2α(1+tanα)21+tan2α-cos2α =1-cos2α =sin2α

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$\frac{1 + \sin 2 α}{\cos (2 α - 2 π) \tan (α - \frac{3 π}{4})} - \frac{1}{4} \sin 2 α [\cot \frac{α}{2} + \cot (\frac{3 π}{2} + \frac{α}{2})] =$

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$\sin^{2} \frac{α}{2}$

$S i n^{2} α$

answer is D.

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Detailed Solution

$\begin{array}{rcl} \frac{1 + \sin 2 α}{\cos (2 α - 2 π) \tan (α - \frac{3 π}{4})} - \frac{1}{4} \sin 2 α [\cot \frac{α}{2} + \cot (\frac{3 π}{2} + \frac{α}{2})] \\ = & \frac{1 + \sin 2 α}{\cos 2 α (\frac{1 + \tan α}{1 - \tan α}) - \frac{1}{4} \sin 2 α (\cot \frac{α}{2} - \tan \frac{α}{2})} \\ = & \frac{1 + \sin 2 α}{(\frac{1 - \tan^{2} α}{1 + \tan^{2} α}) (\frac{1 + tanα}{1 - \tan α})} - \frac{1}{4} 2 \sin α \cos α 2 \cot α \\ = & \frac{1 + \sin 2 α}{\frac{{(1 + \tan α)}^{2}}{1 + \tan^{2} α}} - \cos^{2} α \end{array}$

$\begin{array}{rcl} = & \frac{1 + \frac{2 \tan α}{1 + \tan^{2} α}}{\frac{{(1 + \tan α)}^{2}}{1 + \tan^{2} α}} - \cos^{2} α \\ = & \frac{\frac{1 + \tan^{2} α + 2 \tan α}{1 + \tan^{2} α}}{\frac{{(1 + tanα)}^{2}}{1 + \tan^{2} α}} - \cos^{2} α \\ = & 1 - \cos^{2} α \\ = & \sin^{2} α \end{array}$