Question

The value of $\frac{2{\cos }^2\alpha -1}{2\tan (\frac{\pi }{4}-\alpha ).{\sin }^2(\frac{\pi }{4}+\alpha )}$ is

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (B)

$1$

$\frac{2{\cos }^2\alpha -1}{2\tan (\frac{\pi }{4}-\alpha ){\sin }^2(\frac{\pi }{4}+\alpha )}$

$=\frac{{\cos }^2\alpha -{\sin }^2\alpha }{2\frac{1-\tan \alpha }{1+\tan \alpha }{(\sin \alpha \cos \frac{\pi }{4}+\cos \alpha \sin \frac{\pi }{4})}^2}$

$=\frac{{\cos }^2\alpha -{\sin }^2\alpha }{2\frac{\cos \alpha -\sin \alpha }{\cos \alpha +\sin \alpha }{(\frac{1}{\sqrt{2}}\sin \alpha +\frac{1}{\sqrt{2}}\cos \alpha )}^2}$

$=\frac{{\cos }^2\alpha -{\sin }^2\alpha }{2\frac{{\cos }^2\alpha -{\sin }^2\alpha }{(\cos \alpha +\sin \alpha {)}^2}\frac{1}{2}(\cos \alpha +\sin \alpha {)}^2}$

$=\frac{{\cos }^2\alpha -{\sin }^2\alpha }{{\cos }^2\alpha -{\sin }^2\alpha }=1$