Question

The value of $\frac{1-{\tan }^2(\frac{\pi }{4}-A)}{1+{\tan }^2(\frac{\pi }{4}-A)}$ is
where $A\in [0,\frac{\pi }{2}]$

• A. $\cos 2A$
• B. $\tan 2A$
• C. $\sin 2A$
• D. $\cot 2A$

Answer 1

The correct option is C $\sin 2A$

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