The value of the expression $\begin{aligned} & \frac{1+\sin 2 \alpha}{\cos (2 \alpha-2 \pi) \tan…

The value of the expression $\begin{aligned} & \frac{1+\sin 2 \alpha}{\cos (2 \alpha-2 \pi) \tan \left(\alpha-\frac{3 \pi}{4}\right)} \\ & -\frac{1}{4} \sin 2 \alpha\left[\cot \frac{\alpha}{2}+\cot \left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)\right] \end{aligned}$
  1. $0$
  2. $1$
  3. $\sin ^2 \frac{\alpha}{2}$
  4. $\sin ^2 \alpha$

Solution

Given expression, $\begin{aligned} & \frac{1+\sin 2 \alpha}{\cos (2 \alpha-2 \pi) \tan \left(\alpha-\frac{3 \pi}{4}\right)} \\ & -\frac{1}{4} \sin 2 \alpha\left[\cot \frac{\alpha}{2}+\cot \left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)\right] \\ & \Rightarrow \frac{(\cos \alpha+\sin \alpha)^2}{\cos 2 \alpha\left(\frac{\tan \alpha-\tan \frac{3 \pi}{4}}{1+\tan \alpha \tan \frac{3 \pi}{4}}\right)} \\ & -\frac{1}{4} 2 \sin \alpha \cos \alpha\left(\cot \frac{\alpha}{2}-\tan \frac{\alpha}{2}\right) \\ & =\frac{(\cos \alpha+\sin \alpha)^2}{\cos ^2 \alpha-\sin ^2 \alpha\left(\frac{\tan \alpha+1}{1-\tan \alpha}\right)} \\ & -\frac{1}{4} 2 \sin \alpha \cos \alpha\left(\frac{\cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}}-\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}\right) \\ & =\frac{(\cos \alpha+\sin \alpha) \times(\cos \alpha-\sin \alpha)}{(\cos \alpha-\sin \alpha) \times(\sin \alpha+\cos \alpha)} \\ & -\frac{1}{2} \sin \alpha \cos \alpha\left(\frac{\cos ^2 \frac{\alpha}{2}-\sin ^2 \frac{\alpha}{2}}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}\right) \\ & =1-\frac{\sin \alpha \cos ^2 \alpha}{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}=1-\frac{\sin \alpha \cos ^2 \alpha}{\sin \alpha} \\ & =1-\cos ^2 \alpha=\sin ^2 \alpha \\ & \end{aligned}$

Asked in: AP EAMCET 2016

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